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Counting equally likely outcomes

A STORY

Forty chits in a box

Dhruv holds a wooden box of paper chits steady on the desk. Bhavisha reaches in to draw one.

“There are $40$ chits, one for each of us,” Dhruv says. “The name that comes out reads the notice at assembly.”

“Every chit is cut the same size,” Bhavisha says. “So no name is easier to reach than any other.”

“Then what is the chance the chit has your name on it?” Dhruv asks.

“One chit out of $40$,” Bhavisha says. “So $1/40$.”

“And the chance it has one of our names on it?”

“All $40$ out of $40$. That is $1$. It cannot miss.”

“And the chance it says the headmaster’s name?” Dhruv asks, grinning.

“None of the $40$,” Bhavisha says. “That is $0$. No chance can be less than that, and none can be more than $1$.”

Dhruv tilts the box towards her. “Go on, then. Draw.”

You will learn to count the outcomes that are equally likely, and to write every chance as a number from $0$ to $1$.

A coin comes up heads or tails. A die shows one of six faces. A card drawn from a deck is one of fifty-two.

Toss a coin in your head. List every result it could give. You should find two: heads and tails. Nothing about the coin favours one side over the other.

That is what makes the count useful. When every outcome is equally likely this way, the probability of an event is one count: favourable outcomes over total outcomes. That number is always between $0$ and $1$, inclusive. It is never negative. It is never bigger than one whole.

Two coins, two dice, and a bag of coloured marbles all reduce to the same count. We list every outcome first, then we divide. We check the list against the count each worked example gives.

Probability is a share, not a count: doubling both the favourable and the total outcomes leaves it unchanged.
Dignaga
A SCHOLAR INDIA REMEMBERS

Many thinkers before Dignaga accepted what a trusted person said as a way of knowing. They also accepted a comparison with something similar. Dignaga did not. He kept only two ways: what you see, and what you work out from it. In probability we work the same way. We do not trust a feeling about a coin. We list the outcomes, and we count them.

Dignaga · the kingfisher
Check yourself
  1. This chapter’s opening rule says the probability of an event, once every outcome is equally likely, is
    1. the number of favourable outcomes divided by the number of unfavourable outcomes
    2. the number of all possible outcomes divided by the number of favourable outcomes
    3. the number of favourable outcomes multiplied by the total number of outcomes
    4. the number of favourable outcomes divided by the number of all possible outcomes
    Check your answer
    1. the number of favourable outcomes divided by the number of unfavourable outcomes — The denominator is the TOTAL number of outcomes, not just the unfavourable ones.
    2. the number of all possible outcomes divided by the number of favourable outcomes — The favourable count belongs on top — inverting the fraction gives a number that is not even a valid probability most of the time.
    3. the number of favourable outcomes multiplied by the total number of outcomes — The rule divides the two counts — multiplying them gives a number that is not a probability at all.
    4. ✓ the number of favourable outcomes divided by the number of all possible outcomes — (D) The rule divides the favourable-outcome count by the count of all possible outcomes.
  2. Why must the outcomes being counted be equally likely before this rule can be used?
    1. it assumes every outcome has the same chance
    2. outcomes that are not equally likely cannot be counted or listed at all
    3. equally likely outcomes always come in an even number, like $2$, $4$, or $6$
    4. the rule only works for probabilities greater than one half
    Check your answer
    1. ✓ it assumes every outcome has the same chance — (A) The rule silently assumes each counted outcome has the same chance — it only works when that is true.
    2. outcomes that are not equally likely cannot be counted or listed at all — Unequal outcomes can be listed just fine — the rule still runs, it simply gives a wrong number.
    3. equally likely outcomes always come in an even number, like $2$, $4$, or $6$ — There is no such requirement — a set of equally likely outcomes can be any size, odd or even.
    4. the rule only works for probabilities greater than one half — The rule applies across the whole range from $0$ to $1$ — there is no restriction to values above one half.
  3. The chapter later states a bounds rule, a complement rule, and an elementary-events rule. How are all three connected to this opening frame?
    1. only the bounds rule traces back to the frame; the other two use a separate formula
    2. none of them trace back to the frame; each is its own independent rule
    3. only the elementary-events rule traces back to the frame, since it also mentions outcomes
    4. all three are derived by counting favourable and total outcomes in different situations
    Check your answer
    1. only the bounds rule traces back to the frame; the other two use a separate formula — The complement rule and the elementary-events rule also come from the same favourable-over-total count, not a separate formula.
    2. none of them trace back to the frame; each is its own independent rule — Every later rule in this chapter is built on the same counting idea the frame states — none of them stand apart from it.
    3. only the elementary-events rule traces back to the frame, since it also mentions outcomes — The bounds rule and the complement rule are built from this same counting idea — mentioning “outcomes” is not what makes the link.
    4. ✓ all three are derived by counting favourable and total outcomes in different situations — (D) Each later rule is just the opening favourable-over-total count, applied to a different situation.

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Before you start

You already found the probability of a coin or a die. Now count outcomes from two objects. Try each check below.

If any of these felt new, read the page named before going on.

A tree branches into 2 paths, then 3 more from each, giving 2 times 3 equals 6 outcomes.

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Trials, experiments, and events

●KEY-TERM

A random experiment is a trial with an uncertain result. Every result it could give is known in advance, even though the result itself is not.

Tossing a coin is one example. Nobody can say ahead of time whether it lands heads or tails. But only those two results are possible, and that much is known before the coin is even tossed.

An event is one or more of an experiment’s possible results. Getting a head, when a coin is tossed, is one such event.

Name every possible result first. Only then does calling something an event make sense. We can try this on a die throw: we list all six results first, then check the list against what a die can show.

Your turn: a spinner has four equal sectors, coloured red, blue, green and yellow. List every result one spin can give. (Answer: four results, red, blue, green and yellow, since the spinner has exactly those four sectors and nothing favours one over another.)

The box’s eight chits are all known before anything is drawn, and an event is simply some of those eight picked out.
Check yourself
  1. A random experiment is best described as a trial where
    1. the result is completely unpredictable, and even the possible results are unknown
    2. the result is always the same, no matter how many times the trial is repeated
    3. the result depends on how carefully the trial is carried out
    4. the possible results are known, but which one occurs is not
    Check your answer
    1. the result is completely unpredictable, and even the possible results are unknown — The possible results ARE known in advance — a coin toss can only give a head or a tail. Only which one occurs is uncertain.
    2. the result is always the same, no matter how many times the trial is repeated — An always-the-same result has no uncertainty in it — that rules it out as a random experiment.
    3. the result depends on how carefully the trial is carried out — Care or skill plays no role here — a coin toss or a die throw is uncertain regardless of how carefully it is done.
    4. ✓ the possible results are known, but which one occurs is not — (D) The result is uncertain beforehand, even though the full list of possible results is known.
  2. A student says: “flipping a coin is a random experiment because we can never know anything about it in advance.” What is wrong with this?
    1. the outcomes are known in advance — only which one occurs isn’t
    2. nothing is wrong — a random experiment truly tells us nothing in advance
    3. the coin must be flipped many times before it counts as a random experiment
    4. a random experiment only applies to games of chance, not everyday actions
    Check your answer
    1. ✓ the outcomes are known in advance — only which one occurs isn’t — (A) We already know the two possible outcomes in advance — only which one occurs is uncertain.
    2. nothing is wrong — a random experiment truly tells us nothing in advance — The possible outcomes ARE known in advance — head or tail. It is only which one occurs that stays uncertain.
    3. the coin must be flipped many times before it counts as a random experiment — A single flip already counts as a random experiment — repeating it has nothing to do with the definition.
    4. a random experiment only applies to games of chance, not everyday actions — The definition covers any such trial, not only games of chance — a coin flip qualifies either way.
  3. An event, as this chapter uses the word, is defined as
    1. one or more of the possible results of a trial
    2. the single most likely result of a trial
    3. the total number of possible results a trial can give
    4. any result that has already occurred in a past trial
    Check your answer
    1. ✓ one or more of the possible results of a trial — (A) An event is any one or more of a trial’s possible results, grouped together.
    2. the single most likely result of a trial — An event can be any grouping of possible results, not only the most likely one — a rare outcome is still an event.
    3. the total number of possible results a trial can give — That count is the sample space’s size, not an event — an event is one or more of those results, not their total count.
    4. any result that has already occurred in a past trial — An event is defined from the trial’s possible results, decided before the trial runs — it has nothing to do with a past record.

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Equally likely outcomes

●KEY-TERM

Outcomes are equally likely when nothing favours one over another. A fair coin’s head and tail are equally likely for this reason: nothing about a fair coin favours one side over the other.

Not every experiment keeps this property at every level. Picture a bag holding more red balls than blue ones. Drawing any one ball is equally likely for every ball in the bag. But drawing a red ball is not equally likely to drawing a blue one.

We count the balls first: more reds than blues means the colours cannot be equally likely, even though each single ball still is.

Being equally likely does not automatically pass from individual outcomes to the categories built out of them. We check which level is being counted: individual items, or groups of them.

Your turn: a bag holds $2$ red balls and $5$ blue balls. Is red as likely as blue? (Answer: no, since each of the $7$ balls is equally likely, but $5$ are blue and only $2$ are red.)

The same four names are drawn twice, and only the draw that gives every name one ticket makes the four outcomes equally likely.
Dignaga
A SCHOLAR INDIA REMEMBERS

What does equally likely mean, exactly? Say it without the word likely. Each outcome has the same chance as every other outcome. A fair die has $6$ faces, and each face has the same chance of coming up. A die with one heavy corner does not.

Dignaga · the kingfisher
Check yourself
  1. Outcomes of a trial are called equally likely when
    1. there are exactly two possible outcomes to choose between
    2. no outcome is favoured over another
    3. each outcome has occurred the same number of times so far
    4. the outcomes are listed in a fixed, agreed order
    Check your answer
    1. there are exactly two possible outcomes to choose between — Having exactly two outcomes says nothing about their chances — a weighted coin also has two outcomes, but they are not equally likely.
    2. ✓ no outcome is favoured over another — (B) “Equally likely” means no outcome is favoured over any other by anything in the experiment.
    3. each outcome has occurred the same number of times so far — Equally likely is about the trial itself, decided in advance — not about how many times each outcome has been observed so far.
    4. the outcomes are listed in a fixed, agreed order — The order in which outcomes are written down has nothing to do with whether they are equally likely.
  2. A bag has $5$ red balls and $1$ blue ball. A student says the red and blue outcomes are equally likely, since there are only two colours. What is wrong with this?
    1. nothing is wrong — with two colours, each has probability $1/2$
    2. the outcomes can never be called equally likely once there is more than one ball of any colour
    3. the bag needs equal numbers of every colour before probability can even be discussed
    4. only the six balls are equally likely, not the two colours
    Check your answer
    1. nothing is wrong — with two colours, each has probability $1/2$ — The two colours are not equally likely — red has five times as many balls as blue, so treating them as $1/2$ each is wrong.
    2. the outcomes can never be called equally likely once there is more than one ball of any colour — The six individual balls ARE equally likely — the mistake here is only in treating the two colours as equally likely.
    3. the bag needs equal numbers of every colour before probability can even be discussed — Probability applies to any bag, equal colours or not — an unequal split just means the colours themselves are not equally likely.
    4. ✓ only the six balls are equally likely, not the two colours — (D) The two colours are not equally likely — only the six individual balls are, since red outnumbers blue five to one.
  3. Why is a fair coin’s head and tail equally likely, but a weighted die’s six faces are not?
    1. because a coin has fewer faces than a die
    2. a coin favours neither face; a weighted die does
    3. because a die is thrown while a coin is tossed
    4. because every die is always weighted, unlike every coin
    Check your answer
    1. because a coin has fewer faces than a die — How many faces an object has does not decide fairness — a die with six faces can still be perfectly fair.
    2. ✓ a coin favours neither face; a weighted die does — (B) A fair coin favours neither face, while a weighted die is built to favour some faces over others.
    3. because a die is thrown while a coin is tossed — Tossing versus throwing is just a difference in words for the action — it has no bearing on whether the outcomes are equally likely.
    4. because every die is always weighted, unlike every coin — Most dice in this chapter’s problems are fair — “weighted” describes a specific, unusual die, not every die.

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Count outcomes, not categories

▲CONCEPT
Both bags hold six equally likely balls, but the colours are equally likely only in the bag where every colour has the same count.

The classical probability formula needs equally likely things to count. Falling into equally many categories is not enough.

Picture a bag of coloured balls. It has as many colour-categories as colours. But the colours are equally likely in only one case: when every colour holds exactly the same number of balls. We count the individual balls of each colour first, before deciding anything about the colours themselves.

The individual balls are what is actually equally likely, not the colours. Each ball is as likely to be drawn as any other. Each colour has an equal share of the probability only when every colour holds exactly the same count. With $2$ red balls and $1$ blue ball, is red really as likely as blue?

Count the individual, equally likely items first. We group them into categories only afterwards.

Your turn: a box holds $1$ black pen and $3$ blue pens. Find $P(\text{blue})$. (Answer: $3/4$, not $1/2$, since $3$ of the $4$ pens are blue, and only the blue-against-black split counts, not the number of colours.)

Check yourself
  1. A bag holds $4$ green and $2$ yellow discs, and one is drawn at random. Which statement is correct?
    1. the six discs are equally likely; the colours are not
    2. the two colours are equally likely, since there are only two colours to choose from
    3. the six discs are not equally likely, since some are green and some are yellow
    4. neither the discs nor the colours are equally likely until the bag is shaken
    Check your answer
    1. ✓ the six discs are equally likely; the colours are not — (A) The six discs are equally likely; the two colours are not, since green outnumbers yellow.
    2. the two colours are equally likely, since there are only two colours to choose from — Two categories does not mean equal chances — green has four discs against yellow’s two.
    3. the six discs are not equally likely, since some are green and some are yellow — Each of the six discs has the same chance of being drawn — their colour does not change that.
    4. neither the discs nor the colours are equally likely until the bag is shaken — Shaking the bag plays no role here — the six discs are equally likely by the setup itself, colours are not.
  2. Why does treating the two colours as equally likely give the wrong probability in the discs example?
    1. it assumes equal colour counts, which is false here
    2. because colours cannot be assigned a probability at all
    3. because there is no formula for the probability of drawing a colour
    4. because a disc’s colour changes depending on which one is drawn
    Check your answer
    1. ✓ it assumes equal colour counts, which is false here — (A) Treating colours as equally likely quietly assumes equal counts, which the four-and-two split contradicts.
    2. because colours cannot be assigned a probability at all — A colour is a valid event — $P(\text{green}) = 4/6$ is a real probability, just not $1/2$.
    3. because there is no formula for the probability of drawing a colour — The same favourable-over-total formula applies to a colour event as to any other — it is not a special case needing its own formula.
    4. because a disc’s colour changes depending on which one is drawn — A disc’s colour is fixed before it is drawn — drawing does not change what colour it is.
  3. No matter how many categories a situation seems to offer, what must always be true of the individual outcomes counted in the classical probability formula?
    1. there must be exactly the same number of categories as outcomes
    2. there must be at least two categories to compare
    3. the individual outcomes must be equally likely
    4. the categories must be listed in the same order they appear in the bag
    Check your answer
    1. there must be exactly the same number of categories as outcomes — Matching the number of categories to outcomes is not the requirement — what matters is that the underlying outcomes are equally likely.
    2. there must be at least two categories to compare — A minimum number of categories is not required — the formula works for a single event compared against all outcomes.
    3. ✓ the individual outcomes must be equally likely — (C) The individual outcomes counted must themselves be equally likely, whatever category they are grouped into.
    4. the categories must be listed in the same order they appear in the bag — The order categories are listed in makes no difference — what matters is whether the underlying outcomes are equally likely.
  4. A box has $3$ black pens and $5$ blue pens. One pen is drawn at random. What is $P(\text{black})$?
    1. $1/2$
    2. $3/8$
    3. $3/5$
    4. $5/8$
    Check your answer
    1. $1/2$ — Black and blue are not equally likely categories — the eight individual pens are, giving $P(\text{black}) = 3/8$, not $1/2$.
    2. ✓ $3/8$ — (B) The eight individual pens are equally likely, not the two colours, so $P(\text{black}) = 3/8$.
    3. $3/5$ — This compares black pens to blue pens directly — the formula compares favourable outcomes to ALL outcomes, giving $3/8$.
    4. $5/8$ — This counts the blue pens as favourable instead of black — $P(\text{black}) = 3/8$.

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The classical probability formula

▲CONCEPT
The formula applies to the first three counts, whose outcomes are equally likely, but not to the fourth, the same bag counted by colour.

The probability of an event $E$ is written $P(E)$. It is the number of outcomes favourable to $E$, divided by the number of all possible outcomes.

This formula holds only when every outcome is equally likely. Without that condition, the same division counts nothing meaningful.

A fair die’s six faces meet this condition. So do a fair coin’s two sides. So does a well-shuffled deck’s fifty-two cards. Once the outcomes are confirmed equally likely, we divide favourable by total.

The same one formula applies every time: favourable outcomes, divided by all possible outcomes, once they are confirmed equally likely.

Your turn: five letter cards spell APPLE. Find $P(\text{P})$. (Answer: $2/5$, since $2$ of the $5$ cards, the two Ps, are favourable, out of $5$ cards in all.)

The word has four different letters but five cards, and the formula counts cards: two favourable over five in all.
Check yourself
  1. $P(E)$ is defined as
    1. $(\text{number of all possible outcomes})/(\text{number of outcomes favourable to } E)$
    2. $(\text{number of outcomes favourable to } E)/(\text{number of all possible outcomes})$
    3. $(\text{number of outcomes favourable to } E)/(\text{number of outcomes NOT favourable to } E)$
    4. $(\text{number of outcomes favourable to } E) - (\text{number of all possible outcomes})$
    Check your answer
    1. $(\text{number of all possible outcomes})/(\text{number of outcomes favourable to } E)$ — Inverting the fraction swaps which count is on top — the favourable count belongs in the numerator.
    2. ✓ $(\text{number of outcomes favourable to } E)/(\text{number of all possible outcomes})$ — (B) $P(E)$ divides the count of outcomes favourable to $E$ by the count of all possible outcomes.
    3. $(\text{number of outcomes favourable to } E)/(\text{number of outcomes NOT favourable to } E)$ — The denominator must be the TOTAL number of outcomes, not just the outcomes that are not favourable.
    4. $(\text{number of outcomes favourable to } E) - (\text{number of all possible outcomes})$ — The formula divides the two counts — subtracting them gives a number that is not a probability at all.
  2. This formula only gives a correct probability under one condition. What is it?
    1. the event $E$ must have exactly one favourable outcome
    2. the total number of outcomes must be an even number
    3. every outcome of the experiment must be equally likely
    4. the experiment must be repeated a few times first
    Check your answer
    1. the event $E$ must have exactly one favourable outcome — The formula works for events with any number of favourable outcomes — restricting it to exactly one is not the real condition.
    2. the total number of outcomes must be an even number — There is no requirement on the total being even — the real condition is that the outcomes are equally likely, whatever their count.
    3. ✓ every outcome of the experiment must be equally likely — (C) The formula only holds once every outcome being counted is equally likely.
    4. the experiment must be repeated a few times first — The formula needs no repeated trials — it is computed directly from the outcome counts, before the experiment is even run.
  3. A spinner has $8$ equally likely sectors, numbered $1$ to $8$. What is $P(\text{a number greater than } 5)$?
    1. $5/8$
    2. $3/5$
    3. $1/3$
    4. $3/8$
    Check your answer
    1. $5/8$ — This counts sectors $1$ through $5$ — the question asks for greater than $5$, which is $6$, $7$, $8$, giving $3/8$.
    2. $3/5$ — The denominator must be all $8$ sectors, not $5$ — $3/8$ is correct, not $3/5$.
    3. $1/3$ — There are three favourable sectors, $6$, $7$, $8$, not one — the correct probability is $3/8$.
    4. ✓ $3/8$ — (D) Three sectors, $6$, $7$, $8$, are greater than $5$, out of $8$ in all, giving $3/8$.
  4. One letter is chosen at random from the word “PROBABILITY”. What is the probability that it is a vowel?
    1. $3/11$
    2. $4/12$
    3. $7/11$
    4. $4/11$
    Check your answer
    1. $3/11$ — The word has two $I$’s, both vowels — counting distinct letters instead of positions misses one, giving $3/11$ instead of $4/11$.
    2. $4/12$ — “PROBABILITY” has $11$ letters, not $12$ — recount them: P-R-O-B-A-B-I-L-I-T-Y.
    3. $7/11$ — This counts the consonants ($11 - 4 = 7$) — the question asks for vowels, which is $4/11$.
    4. ✓ $4/11$ — (D) “PROBABILITY” has $11$ letters, of which $O$, $A$, $I$, $I$ are the $4$ vowels, giving $4/11$.

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Elementary events sum to one

▲CONCEPT

An event with exactly one outcome is called an elementary event. Getting a head on a coin toss is one. So is getting a $3$ on a die throw.

Every experiment breaks down into its full set of elementary events, one for each possible outcome. Take a die: it has six elementary events, one for each face.

*The probabilities of all the elementary events of an experiment always add up to $1$.* We add the die’s six probabilities together, $1/6$ six times over, and land on exactly $1$.

That total is a check worth running on any finished list of probabilities. If the elementary events do not add to $1$, something in the counting has gone wrong.

Your turn: a spinner has three equal sectors, numbered $1$, $2$ and $3$. Find each sector’s probability, and check they add to $1$. (Answer: each sector is $1/3$, and $1/3 + 1/3 + 1/3 = 1$.)

A coin’s strip splits into two equal pieces and a die’s strip into six, one piece for each equally likely outcome, together making one whole.
Check yourself
  1. An event with exactly one outcome, such as “getting a $3$” on a die throw, is called
    1. a compound event
    2. an impossible event
    3. an elementary event
    4. a complementary event
    Check your answer
    1. a compound event — A compound event covers more than one outcome — an event with exactly one outcome is an elementary event instead.
    2. an impossible event — An impossible event never happens at all — “getting a $3$” is a perfectly possible single outcome, so it is an elementary event.
    3. ✓ an elementary event — (C) An event built from exactly one outcome is called an elementary event.
    4. a complementary event — A complementary event is defined against another event (“not $E$”) — having exactly one outcome is what makes an event elementary.
  2. Why must the probabilities of all the elementary events of an experiment always add up to $1$?
    1. because there are always exactly two elementary events in any experiment
    2. because each elementary event must have probability exactly $1/2$
    3. the elementary events cover every outcome exactly once
    4. because probability is measured out of a total of $1$ percent
    Check your answer
    1. because there are always exactly two elementary events in any experiment — The number of elementary events varies by experiment — a die has six, not two — so this is not the reason.
    2. because each elementary event must have probability exactly $1/2$ — Only a two-outcome experiment gives $1/2$ each — a die’s six elementary events each have probability $1/6$, and they still sum to $1$.
    3. ✓ the elementary events cover every outcome exactly once — (C) The elementary events together cover every outcome exactly once, so their counts add to the total, giving a sum of $1$.
    4. because probability is measured out of a total of $1$ percent — Probability is a plain number between $0$ and $1$, not a percentage — the sum-to-one comes from the outcomes covering every case, not from a percent scale.
  3. A fair die is thrown once. What is the sum of the probabilities of the six elementary events, “getting a $1$” through “getting a $6$”?
    1. $1/6$
    2. $6$
    3. $1$
    4. $3.5$
    Check your answer
    1. $1/6$ — $1/6$ is the probability of just ONE elementary event — the question asks for the sum of all six, which is $1$.
    2. $6$ — Adding the face numbers $1$ through $6$ gives $21$, not $6$ — and either way, the question asks for the sum of PROBABILITIES, which is $1$.
    3. ✓ $1$ — (C) Each of the six elementary events has probability $1/6$, and $6 \cdot 1/6 = 1$.
    4. $3.5$ — Averaging the face numbers gives the die’s expected value, a different idea — the sum of the six probabilities is $1$.

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Between zero and one, always

▲CONCEPT

The probability of any event $E$ satisfies $0 \leq P(E) \leq 1$. It can never fall below zero, and it can never rise above one whole. We give both ends of that range their own name.

An event that can never happen is called an impossible event, and it carries $P(E) = 0$. An event that must always happen is called a sure event, or a certain event, and it carries $P(E) = 1$.

No genuine probability is ever negative or bigger than one whole.

Your turn: someone claims an event has probability $5/4$. What went wrong? (Answer: more favourable outcomes were counted than there are outcomes in all, since no probability can ever be bigger than $1$.)

Three rows shade the same six die faces for three events, from no faces shaded to all six.
Dignaga
A SCHOLAR INDIA REMEMBERS

A friend says the chance of rain is $1.5$. Do you need to see the working? No. You can reject it at once. A probability counts favourable outcomes out of all the outcomes. The part cannot be bigger than the whole. So every probability is between $0$ and $1$, or equal to one of them.

Dignaga · the kingfisher
Check yourself
  1. For any event $E$, $P(E)$ always satisfies
    1. $0 < P(E) < 1$
    2. $-1 \leq P(E) \leq 1$
    3. $0 \leq P(E) \leq 100$
    4. $0 \leq P(E) \leq 1$
    Check your answer
    1. $0 < P(E) < 1$ — This excludes $0$ and $1$ themselves — an impossible event has $P(E) = 0$ and a sure event has $P(E) = 1$, so both endpoints must be included.
    2. $-1 \leq P(E) \leq 1$ — A probability can never be negative — the lower bound is $0$, not $-1$.
    3. $0 \leq P(E) \leq 100$ — Probability is measured from $0$ to $1$ — the $0$-to-$100$ range belongs to percentages, a different scale.
    4. ✓ $0 \leq P(E) \leq 1$ — (D) Every probability lies between $0$ and $1$, both values included.
  2. A student calculates a probability of $1.2$ for an event. What does this tell you?
    1. nothing is wrong — probabilities above $1$ are possible for rare events
    2. the event must be a sure event, since $1.2$ rounds to $1$
    3. the event is impossible, since its probability lies outside the normal range
    4. a mistake was made somewhere, since probability can never exceed $1$
    Check your answer
    1. nothing is wrong — probabilities above $1$ are possible for rare events — No probability can ever exceed $1$, however rare the event — a value of $1.2$ always signals a counting mistake.
    2. the event must be a sure event, since $1.2$ rounds to $1$ — Rounding $1.2$ to $1$ does not make it valid — a probability calculation should never reach a value above $1$ in the first place.
    3. the event is impossible, since its probability lies outside the normal range — An impossible event has probability $0$, not a value above $1$ — a result of $1.2$ signals a counting mistake, not impossibility.
    4. ✓ a mistake was made somewhere, since probability can never exceed $1$ — (D) A probability of $1.2$ signals a mistake, since no probability can exceed $1$.
  3. Why can an impossible event’s probability never be less than $0$, even though it never happens?
    1. zero favourable outcomes over the total is simply $0$
    2. because probability is only measured for events that could still happen
    3. because an impossible event technically has one favourable outcome that never gets picked
    4. because the total number of outcomes always cancels out to a positive number
    Check your answer
    1. ✓ zero favourable outcomes over the total is simply $0$ — (A) An impossible event has $0$ favourable outcomes, and $0$ divided by any total is $0$, never negative.
    2. because probability is only measured for events that could still happen — An impossible event is still a valid event — the formula still applies to it and correctly gives $0$.
    3. because an impossible event technically has one favourable outcome that never gets picked — An impossible event has zero favourable outcomes by definition — there is no hidden favourable outcome waiting to be picked.
    4. because the total number of outcomes always cancels out to a positive number — Nothing cancels here — $0$ favourable outcomes divided by any positive total is simply $0$.
  4. A bag has only red balls. What is the probability of drawing a blue ball?
    1. it cannot be found without knowing how many red balls there are
    2. $1$, since blue is the only colour missing
    3. $0$
    4. undefined, since there are no blue balls to count
    Check your answer
    1. it cannot be found without knowing how many red balls there are — The number of red balls does not matter here — there are $0$ blue balls to draw, whatever the total, so the probability is $0$.
    2. $1$, since blue is the only colour missing — A missing colour gives probability $0$, not $1$ — $1$ describes a sure event, the opposite of an impossible one.
    3. ✓ $0$ — (C) Drawing a blue ball is an impossible event here, so its probability is $0$.
    4. undefined, since there are no blue balls to count — This is perfectly defined, not undefined — $0$ favourable outcomes over any total still gives a valid probability, $0$.

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Complementary events

▲CONCEPT

For any event $E$, the event ‘not $E$’ is called the complement of $E$. It happens exactly when $E$ does not, and one of the two always does.

The two probabilities always add to one. $P(E) + P(\text{not E}) = 1$, so $P(\text{not E}) = 1 - P(E)$. We find the other by one subtraction, once either probability is known.

Finding $P(\text{not E})$ this way is often the shorter route, especially when ‘not $E$’ is easier to count than $E$ itself. The subtraction only gives the true complement when ‘not $E$’ really is everything $E$ is not: nothing more, nothing less.

Your turn: $P(E) = 3/8$. Find $P(\text{not E})$. (Answer: $5/8$, since $P(\text{not E}) = 1 - 3/8 = 5/8$.)

A circle marks event E inside a rectangle, and the rest of the rectangle marks its complement, together filling the whole shape.
Check yourself
  1. For any event $E$, the complement “not $E$” is the event that
    1. happens at the same time as $E$, but with a different outcome
    2. happens exactly when $E$ does not happen
    3. always has exactly the same probability as $E$
    4. can never actually occur in a real trial
    Check your answer
    1. happens at the same time as $E$, but with a different outcome — $E$ and “not $E$” can never both happen at once — one occurs exactly when the other does not.
    2. ✓ happens exactly when $E$ does not happen — (B) “Not $E$” happens precisely when $E$ fails to happen, and never at the same time as $E$.
    3. always has exactly the same probability as $E$ — The two probabilities are equal only in the special case of $1/2$ each — in general, $P(E)$ and $P(\text{not} E)$ differ.
    4. can never actually occur in a real trial — “Not $E$” is a real, possible event — it occurs whenever $E$ itself is not a sure event.
  2. Why does $P(E) + P(\text{not} E)$ always equal $1$?
    1. together, $E$ and “not $E$” cover every outcome exactly once
    2. because $E$ and “not $E$” are always equally likely
    3. because every experiment has exactly two possible events in total
    4. because probability values are always written as fractions that must add to one whole
    Check your answer
    1. ✓ together, $E$ and “not $E$” cover every outcome exactly once — (A) Between them, $E$ and “not $E$” cover every possible outcome exactly once, so their counts add up to the total.
    2. because $E$ and “not $E$” are always equally likely — $E$ and “not $E$” are equally likely only when both equal $1/2$ — the sum-to-one rule holds even when they are very unequal.
    3. because every experiment has exactly two possible events in total — An experiment can have many possible events — $E$ and “not $E$” are just one such pair, chosen to cover all outcomes between them.
    4. because probability values are always written as fractions that must add to one whole — The sum-to-one is not a convention about fractions — it comes from $E$ and “not $E$” together covering every outcome exactly once.
  3. $P(E) = 3/7$ for some event $E$. What is $P(\text{not} E)$?
    1. $3/7$
    2. $-3/7$
    3. $4/7$
    4. $1/7$
    Check your answer
    1. $3/7$ — $E$ and “not $E$” need not be equal — applying the complement rule, $1 - 3/7 = 4/7$.
    2. $-3/7$ — The complement rule subtracts $P(E)$ from $1$ — it does not negate it. $1 - 3/7 = 4/7$.
    3. ✓ $4/7$ — (C) $P(\text{not} E) = 1 - 3/7 = 4/7$.
    4. $1/7$ — A careful subtraction gives $1 - 3/7 = 4/7$, not $1/7$.
  4. Rather than count the outcomes favourable to “not an ace” directly, what shortcut does the complement rule give?
    1. compute $P(\text{ace})$ first, and subtract it from $1$
    2. count aces in each suit separately and average the four results
    3. double the probability of drawing a red ace
    4. find the probability of drawing a king instead, since kings and non-aces are the same size
    Check your answer
    1. ✓ compute $P(\text{ace})$ first, and subtract it from $1$ — (A) The shortcut is to find $P(\text{ace})$ first, then subtract it from $1$ to get $P(\text{not ace})$.
    2. count aces in each suit separately and average the four results — There is no per-suit averaging step here — the complement rule only needs $P(\text{ace})$, subtracted from $1$.
    3. double the probability of drawing a red ace — Doubling the red-ace probability has no bearing on “not an ace” — the shortcut is simply $1 - P(\text{ace})$.
    4. find the probability of drawing a king instead, since kings and non-aces are the same size — Kings and non-aces are not the same size at all — the complement of “ace” is found from $P(\text{ace})$ itself, not from a different rank.

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Outcomes from more than one object

▲CONCEPT

Some experiments involve more than one object at once. Two coins tossed together is one example, and two dice thrown together is another.

The outcome of such an experiment is a pair, and which object shows what matters to the count. We list every outcome of tossing two coins before reading on.

Two coins give four equally likely outcomes: $H H$, $H T$, $T H$, $T T$. We check that list against this one: four outcomes, not three. $H T$ and $T H$ sound similar in words, one head and one tail either way. But they are two different outcomes: it matters which coin shows the head.

*For $n$ objects, each with $k$ equally likely outcomes, the combined experiment has $k^n$ equally likely outcomes.* Two dice give $6 \cdot 6 = 36$ ordered pairs, since each die has six faces and there are two dice.

Your turn: two spinners each have three equal sectors, A, B and C. Find how many ordered outcomes one spin of both gives. (Answer: $3^2 = 9$: AA, AB, AC, BA, BB, BC, CA, CB, CC.)

Rolling two dice gives ordered pairs, so the sample space is a six-by-six grid, empty before any event is marked.
On the same grid, the six doubles fall in a straight line along the diagonal, since a double is a cell where both dice agree.
Check yourself
  1. When two coins are tossed together, the equally likely outcomes are
    1. $H H$, $T T$ — two outcomes, since only matching pairs count
    2. $H H$, $H T$, $T H$, $T T$ — four outcomes in all
    3. $H T$, $T H$ — two outcomes, since matching pairs are not really different
    4. $H H$, $H T$, $T H$, $T T$, $H H$ — five outcomes, counting $H H$ once for each coin
    Check your answer
    1. $H H$, $T T$ — two outcomes, since only matching pairs count — The mixed outcomes, $H T$ and $T H$, are equally valid results — dropping them leaves only two of the four.
    2. ✓ $H H$, $H T$, $T H$, $T T$ — four outcomes in all — (B) Two coins tossed together give four equally likely outcomes: $H H$, $H T$, $T H$, $T T$.
    3. $H T$, $T H$ — two outcomes, since matching pairs are not really different — The matching outcomes, $H H$ and $T T$, are equally valid results — dropping them leaves only two of the four.
    4. $H H$, $H T$, $T H$, $T T$, $H H$ — five outcomes, counting $H H$ once for each coin — Each of the four outcomes is counted once — counting $H H$ twice gives five instead of the correct four.
  2. Why does tossing two coins give four outcomes rather than three?
    1. because a coin can secretly land on its edge, adding a fourth case
    2. because probability rules always require an even number of outcomes
    3. which coin shows the head matters — $H T$ and $T H$ differ
    4. because two coins must always be tossed one after the other, never together
    Check your answer
    1. because a coin can secretly land on its edge, adding a fourth case — An edge landing is not part of this chapter’s outcome set — the fourth outcome comes from telling $H T$ and $T H$ apart, not from a coin balancing on its rim.
    2. because probability rules always require an even number of outcomes — There is no rule requiring an even number of outcomes — a single die already has six, an even number by coincidence, not by requirement.
    3. ✓ which coin shows the head matters — $H T$ and $T H$ differ — (C) “One head, one tail” hides two different outcomes, $H T$ and $T H$, since which coin shows the head matters.
    4. because two coins must always be tossed one after the other, never together — Whether the coins are tossed together or one after another makes no difference to the four outcomes — the count comes from telling the coins apart, not from timing.
  3. Two dice thrown together give $36$ equally likely outcomes, from $6 \cdot 6$. Why is it $6 \cdot 6$ rather than $6 + 6 = 12$?
    1. because $6 + 6$ only counts the outcomes where the two dice show different numbers
    2. because $12$ already counts every outcome where both dice show the same number twice
    3. because one die’s outcomes must be squared before adding the other die’s six outcomes
    4. each first-die outcome pairs with all six second-die outcomes
    Check your answer
    1. because $6 + 6$ only counts the outcomes where the two dice show different numbers — $6 + 6$ does not correctly count any real subset of outcomes — the right operation is multiplying, since each pairing is independent.
    2. because $12$ already counts every outcome where both dice show the same number twice — The gap between $12$ and $36$ is not from double-counting matches — it is from adding the two counts instead of multiplying them.
    3. because one die’s outcomes must be squared before adding the other die’s six outcomes — No squaring-then-adding step exists — the two dice’s outcome counts are multiplied directly, giving $6 \cdot 6 = 36$.
    4. ✓ each first-die outcome pairs with all six second-die outcomes — (D) Every one of the first die’s $6$ outcomes pairs with every one of the second die’s $6$ outcomes, giving $6 \cdot 6$.
  4. Two dice are thrown together. How many of the $36$ equally likely outcomes have the first die showing an even number?
    1. $3$
    2. $12$
    3. $6$
    4. $18$
    Check your answer
    1. $3$ — The first die has $3$ even faces, but each one pairs with all $6$ second-die outcomes — $3 \cdot 6 = 18$, not $3$.
    2. $12$ — The first die has three even faces, $2$, $4$, $6$, not two — $3 \cdot 6 = 18$, not $12$.
    3. $6$ — Each even face pairs with all $6$ second-die outcomes, not just one — the true count is $3 \cdot 6 = 18$.
    4. ✓ $18$ — (D) The first die has $3$ even faces, each pairing with all $6$ outcomes of the second die, giving $3 \cdot 6 = 18$.

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Two coins give four outcomes, not three

✕MISCONCEPTION

Toss two coins in your head. List every outcome. Tossing two coins looks like it should give three outcomes: two heads, two tails, or one of each. So surely each of those three has probability $1/3$?

That count is wrong. The four outcomes $H H$, $H T$, $T H$ and $T T$ are all equally likely. Which coin shows the head matters, even though ‘one of each’ sounds like a single description.

*Exactly one head has probability $2/4 = 1/2$, not $1/3$.* Two of the four equally likely outcomes, $H T$ and $T H$, give exactly one head, and we can count them again to check.

Listing three descriptions of an outcome is not the same as listing three equally likely outcomes. Only the full list of all four gives the correct count: $H H$, $H T$, $T H$, $T T$. Check for this trap in every sample space, and count the outcomes yourself from scratch.

Your turn: two coins are tossed. Find $P(\text{two tails})$. (Answer: $1/4$, since only $T T$ of the four equally likely outcomes gives two tails.)

Check yourself
  1. A student says: “Tossing two coins gives three outcomes — two heads, two tails, or one of each — so each has probability $1/3$.” What is wrong with this?
    1. nothing is wrong — each of the three outcomes does have probability $1/3$
    2. the mistake is only in the fraction $1/3$; the three outcomes are listed correctly
    3. the four outcomes are equally likely, giving $1/2$
    4. two coins cannot be tossed together at all — they must be tossed one after another
    Check your answer
    1. nothing is wrong — each of the three outcomes does have probability $1/3$ — There are four equally likely outcomes, not three — treating them as three gives $1/3$ where the true answer is $1/2$.
    2. the mistake is only in the fraction $1/3$; the three outcomes are listed correctly — The three-outcome list is itself the mistake — $H T$ and $T H$ are two separate outcomes, not one “mixed” case, so there are really four.
    3. ✓ the four outcomes are equally likely, giving $1/2$ — (C) The four outcomes $H H$, $H T$, $T H$, $T T$ are equally likely, so exactly one head has probability $2/4 = 1/2$.
    4. two coins cannot be tossed together at all — they must be tossed one after another — Two coins can absolutely be tossed together — the real issue is that “one head, one tail” hides two distinct outcomes, $H T$ and $T H$.
  2. Two coins are tossed. A student lists the outcomes as “both heads”, “both tails”, and “mixed”, assigning each probability $1/3$. What is $P(\text{exactly one head})$ really?
    1. $1/3$, matching the student’s three-outcome count
    2. $1/2$ — “mixed” is really two outcomes, $H T$ and $T H$
    3. $1/4$, since only $H T$ counts as “mixed”, not $T H$
    4. $2/3$, since two of the student’s three listed outcomes are not “both heads”
    Check your answer
    1. $1/3$, matching the student’s three-outcome count — The three-outcome count is the flaw itself — “mixed” is really two outcomes, giving $1/2$, not $1/3$.
    2. ✓ $1/2$ — “mixed” is really two outcomes, $H T$ and $T H$ — (B) “Mixed” hides two separate outcomes, $H T$ and $T H$, giving $P(\text{exactly one head}) = 2/4 = 1/2$.
    3. $1/4$, since only $H T$ counts as “mixed”, not $T H$ — Both $H T$ and $T H$ count as “exactly one head” — dropping $T H$ undercounts, giving $1/4$ instead of the correct $1/2$.
    4. $2/3$, since two of the student’s three listed outcomes are not “both heads” — This still uses the flawed three-outcome list — with the correct four equally likely outcomes, $P(\text{exactly one head}) = 1/2$.
  3. What exactly goes wrong when “one head, one tail” is treated as a single outcome?
    1. it collapses $H T$ and $T H$ into one outcome
    2. nothing goes wrong, since $H T$ and $T H$ describe the exact same physical result
    3. the total number of outcomes becomes too large, not too small
    4. the two heads and two tails outcomes also need to be split into two each
    Check your answer
    1. ✓ it collapses $H T$ and $T H$ into one outcome — (A) Treating “one head, one tail” as one outcome collapses $H T$ and $T H$ together, breaking the equally-likely count.
    2. nothing goes wrong, since $H T$ and $T H$ describe the exact same physical result — $H T$ and $T H$ are different results — which coin lands heads is a real distinction once the two coins are told apart.
    3. the total number of outcomes becomes too large, not too small — Collapsing $H T$ and $T H$ together makes the count SMALLER, three instead of four — not larger.
    4. the two heads and two tails outcomes also need to be split into two each — $H H$ and $T T$ are each already one full outcome — only “one head, one tail” hides two distinct ones, $H T$ and $T H$.
  4. Two coins are tossed. What is $P(\text{at least one head})$?
    1. $2/3$
    2. $1/2$
    3. $3/4$
    4. $1/4$
    Check your answer
    1. $2/3$ — This uses the flawed three-outcome count — with the correct four outcomes, three qualify, giving $3/4$, not $2/3$.
    2. $1/2$ — “At least one head” includes $H H$ as well as $H T$ and $T H$ — that is $3/4$, not the $1/2$ for “exactly one head”.
    3. ✓ $3/4$ — (C) Three of the four outcomes, $H H$, $H T$, $T H$, have at least one head, giving $3/4$.
    4. $1/4$ — “At least one head” is satisfied by $H H$, $H T$, and $T H$ — counting only $H H$ misreads it as “both heads”.
  5. A student throws two dice and lists three outcomes — “first die bigger”, “second die bigger”, “both equal” — claiming each has probability $1/3$. Is this correct?
    1. yes, since exactly one of the three must happen
    2. yes, but only because both dice look identical
    3. no — the $36$ pairs are equally likely, not the three
    4. no, but only because two dice can never show “both equal”
    Check your answer
    1. yes, since exactly one of the three must happen — That exactly one of the three happens is true, but does not make them equally likely — they group unequal numbers of the $36$ pairs.
    2. yes, but only because both dice look identical — Whether the dice look identical changes nothing — the three events are unequal because they group different numbers of the $36$ pairs.
    3. ✓ no — the $36$ pairs are equally likely, not the three — (C) The $36$ ordered pairs are equally likely; the three listed events group unequal numbers of pairs, so none is $1/3$.
    4. no, but only because two dice can never show “both equal” — “Both equal” happens for six of the $36$ pairs, $(1,1)$ through $(6,6)$ — it is a real, possible event, just not equally weighted with the other two.
Grouping two coins’ results into three bands looks fair, but splitting the wide middle band into its two separate outcomes gives the count of four.
Three coins are tossed. What is the probability of exactly two heads?

Weaker. Group the results by how many heads show: three heads, two heads, one head, no heads. Four descriptions, so exactly two heads looks like $1/4$. But ‘two heads’ can happen with the tail on the first coin, on the second, or on the third: three different tosses. ‘Three heads’ can happen in only one way. The four descriptions are not equally likely, so $1/4$ cannot be right.

Stronger. List the $2 \cdot 2 \cdot 2 = 8$ equally likely outcomes: $H H H$, $H H T$, $H T H$, $T H H$, $H T T$, $T H T$, $T T H$, $T T T$. Exactly two heads are $H H T$, $H T H$ and $T H H$, so the probability is $3/8$. The four descriptions cover $1 + 3 + 3 + 1 = 8$ outcomes between them.

Dignaga
A SCHOLAR INDIA REMEMBERS

Two heads, two tails, or one of each. Are those three outcomes of one kind? No. One of each can happen in two ways: HT and TH. So there are four equally likely outcomes. The chance of one head and one tail is $2/4 = 1/2$.

Dignaga · the kingfisher

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Twelve face cards, not sixteen

✕MISCONCEPTION
A grid of playing cards marks only the jack, queen and king as face cards, 12 in all, with the ace excluded.

A deck has four aces, four kings, four queens and four jacks. That looks like sixteen face cards, one card of each rank in every suit.

It is not sixteen. Only the jack, the queen and the king carry a face. Each suit holds three of them, so the deck holds $12$ face cards in all. An ace shows a single suit symbol and no face at all. It is never counted among the face cards.

We can check this suit by suit: the jack, the queen and the king of hearts carry a face, and the ace of hearts does not.

Your turn: how many of the hearts are face cards? (Answer: three, the jack, the queen and the king of hearts. The ace of hearts is not one.)

Siddharth stands at a windowsill, pencil lifted, notebook open, a pencil box and an eraser beside it.
First try

A deck has $4$ aces, $4$ kings, $4$ queens and $4$ jacks. I called all $16$ of them face cards and got $P(\text{not a face card}) = 1 - 16/52 = 36/52$.

Second look

The ace carries a single pip, not a face. Only the king, the queen and the jack carry a face, $3$ in every suit, $12$ in the deck. So $P(\text{not a face card}) = 1 - 12/52 = 40/52 = 10/13$.

Count the ace with the number cards, not with the ones that carry a face.

One card is drawn from a deck. What is the probability it is a red face card?

Weaker. Count an ace, a king, a queen and a jack in each red suit: $8$ red face cards, so $P(\text{red face card}) = 8/52 = 2/13$. But look at the ace of hearts. It carries a single heart and no face, so it cannot be a face card. The count of $8$ is two too many.

Stronger. Only the jack, the queen and the king carry a face, three in each red suit, so there are $3 \cdot 2 = 6$ red face cards and $P(\text{red face card}) = 6/52 = 3/26$. The other $20$ red cards are the ace and $2$ to $10$ in each red suit: $2 \cdot 10 = 20$, and $20 + 6 = 26$.

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Worked: heads and tails

Check yourself
  1. A coin is tossed once. Let $E$ be “getting a head”. $P(E)$ equals
    1. $1$
    2. $0$
    3. $1/2$
    4. $1/4$
    Check your answer
    1. $1$ — A head is not certain — a tail is equally possible, so $P(E) = 1/2$, not $1$.
    2. $0$ — A head is a genuinely possible outcome, not an impossible one — $P(E) = 1/2$, not $0$.
    3. ✓ $1/2$ — (C) One of the coin’s two equally likely outcomes is a head, giving $P(E) = 1/2$.
    4. $1/4$ — A single coin toss has only two outcomes, head and tail, not four — $P(E) = 1/2$.
  2. In a single coin toss, why do $P(\text{head})$ and $P(\text{tail})$ each equal exactly $1/2$?
    1. because a coin toss is always fair by law
    2. because heads and tails are opposite sides of the same coin
    3. because $1/2$ is the only fraction less than $1$
    4. two equally likely outcomes, one each for head and tail
    Check your answer
    1. because a coin toss is always fair by law — There is no external rule making a toss fair — the $1/2$ comes from there being exactly two equally likely outcomes.
    2. because heads and tails are opposite sides of the same coin — Being opposite sides is a fact about the coin’s shape — the $1/2$ split comes from there being two equally likely outcomes, not from the sides being opposite.
    3. because $1/2$ is the only fraction less than $1$ — Many fractions are less than $1$ — the coin gives exactly $1/2$ because it has two equally likely outcomes, one favourable to each event.
    4. ✓ two equally likely outcomes, one each for head and tail — (D) Two equally likely outcomes exist, and each of head and tail claims exactly one of them.
  3. A coin is tossed once. What is $P(\text{head}) + P(\text{tail})$, and why?
    1. $1/2$, since only one of the two can happen at a time
    2. $2$, since two separate probabilities are being added
    3. $1$, the coin’s two elementary events
    4. it cannot be found without tossing the coin first
    Check your answer
    1. $1/2$, since only one of the two can happen at a time — That one outcome happens at a time is true, but the question asks for the SUM of both probabilities, which is $1$, not $1/2$.
    2. $2$, since two separate probabilities are being added — The two probabilities come from splitting the same two outcomes between them — their sum is $1$, not $2$.
    3. ✓ $1$, the coin’s two elementary events — (C) $P(\text{head}) + P(\text{tail}) = 1$, since the two are the coin’s complete set of elementary events.
    4. it cannot be found without tossing the coin first — No toss is needed to find this — the sum of $1/2$ and $1/2$ is computed directly from the outcome count, giving $1$.
Worked example

A coin toss: head and tail

  1. one coin, two equally likely outcomes: $H$, $T$
    We list both possible results first, before naming any event.
  2. $E$ = ‘getting a head’ has $1$ favourable outcome out of $2$: $P(E) = 1/2$
    This is the classical probability formula, applied to the simplest two-outcome experiment.
  3. $F$ = ‘getting a tail’ has $1$ favourable outcome out of $2$: $P(F) = 1/2$
    The same formula, applied to the other elementary event.
  4. $P(E) + P(F) = 1/2 + 1/2 = 1$
    *Check that head and tail add to $1$.* They are the coin’s only two elementary events, so their probabilities must sum to a whole.

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Worked: greater than four, impossible, sure

Check yourself
  1. A fair die is thrown once. What is $P(\text{a number greater than } 4)$?
    1. $1/3$
    2. $2/3$
    3. $1/2$
    4. $4/6$
    Check your answer
    1. ✓ $1/3$ — (A) Two faces, $5$ and $6$, are greater than $4$, out of $6$ in all, giving $2/6 = 1/3$.
    2. $2/3$ — This counts faces $1$ through $4$ — the question asks for greater than $4$, which is only $5$ and $6$, giving $1/3$.
    3. $1/2$ — Only two faces, $5$ and $6$, are greater than $4$ — not three — giving $1/3$, not $1/2$.
    4. $4/6$ — The favourable count comes from how many faces qualify, which is $2$, not from the number $4$ itself.
  2. In the die example, $P(\text{getting} 8) = 0$. What does this illustrate about the classical formula?
    1. no favourable outcomes among the possible ones gives $0$
    2. that the formula fails whenever the desired outcome is a large number
    3. that a six-sided die secretly has an eighth face that is never rolled
    4. that the total number of outcomes must be increased to include $8$
    Check your answer
    1. ✓ no favourable outcomes among the possible ones gives $0$ — (A) An event with zero favourable outcomes, like getting an $8$ on a die, always gives probability $0$.
    2. that the formula fails whenever the desired outcome is a large number — The size of the number is irrelevant — $P = 0$ comes from $8$ never appearing on the die, not from $8$ being a large number.
    3. that a six-sided die secretly has an eighth face that is never rolled — A standard die has exactly six faces — there is no hidden eighth face; $8$ is simply outside the die’s outcome set.
    4. that the total number of outcomes must be increased to include $8$ — The total stays at six, the die’s actual faces — nothing needs to be added to make room for $8$.
  3. For the same die, $P(\text{a number less than } 7)$ is
    1. $6/7$
    2. $5/6$
    3. $0$, mistaking this for the probability of rolling a $7$
    4. $1$
    Check your answer
    1. $6/7$ — The die has six faces, not seven — every one of them is less than $7$, giving $6/6 = 1$, not $6/7$.
    2. $5/6$ — All six faces, $1$ through $6$, are less than $7$ — leaving one out gives $5/6$ instead of the correct $1$.
    3. $0$, mistaking this for the probability of rolling a $7$ — This confuses “less than $7$” with “equal to $7$” — every face qualifies as less than $7$, giving $1$, not $0$.
    4. ✓ $1$ — (D) Every one of the die’s six faces is less than $7$, giving $6/6 = 1$.
  4. Among the three events from the worked example — “greater than $4$”, “getting $8$”, “less than $7$” — which is a sure event?
    1. “getting $8$”, since it names one specific number
    2. “greater than $4$”, since it includes the die’s largest face, $6$
    3. none of them, since a die can never guarantee a sure event
    4. “less than $7$”, since $P = 1$
    Check your answer
    1. “getting $8$”, since it names one specific number — Naming a specific number does not make an event sure — “getting $8$” is in fact the impossible event, with $P = 0$.
    2. “greater than $4$”, since it includes the die’s largest face, $6$ — Including the largest face is not enough — faces $1$ through $4$ still fail to qualify, so this event is neither impossible nor sure.
    3. none of them, since a die can never guarantee a sure event — A die CAN give a sure event — “less than $7$” is satisfied by every face, making $P = 1$.
    4. ✓ “less than $7$”, since $P = 1$ — (D) “Less than $7$” is the sure event, since every face qualifies, giving $P = 1$.
A number line from 0 to 1 marks P(F) = 0 at impossible, P(E) = 1/3, and P(G) = 1 at sure.
Worked example

A die throw: three events classified

  1. one die, six equally likely outcomes: $1, 2, 3, 4, 5, 6$
    We list every outcome first, before deciding which ones satisfy each event.
  2. $E$ = ‘a number greater than $4$’ has outcomes $5, 6$: $P(E) = 2/6 = 1/3$
    $4$ itself does not satisfy ‘greater than $4$’, so only $5$ and $6$ count.
  3. $F$ = ‘getting $8$’ has no favourable outcome: $P(F) = 0/6 = 0$
    This is an impossible event: $8$ is not among the die’s six faces.
  4. $G$ = ‘a number less than $7$’ has outcomes $1$ to $6$, all six: $P(G) = 6/6 = 1$
    This is a sure event: every face of the die is less than $7$.
  5. $0 \leq 0 \leq 1/3 \leq 1 \leq 1$
    *Check that all three answers fall inside $0$ to $1$.* Line up $0$, $1/3$ and $1$: every one sits inside the range a probability must, from impossible up to sure.
Find the probability of any event on one die.
  1. List the six outcomes Write out every result the die can show: $1$, $2$, $3$, $4$, $5$, $6$.
  2. Pick out the favourable ones For the event, list which of those six outcomes satisfy it. ‘Greater than $4$’ picks out $5$ and $6$.
  3. Divide by six Divide the count of favourable outcomes by $6$, the total number of outcomes. ‘Greater than $4$’ gives $2/6 = 1/3$.
  4. Check the range Every answer must sit between $0$ and $1$. An event with no favourable outcome gives $0$, and an event every outcome satisfies gives $1$.

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Worked: counting marbles, not colours

Check yourself
  1. In the marbles example ($3$ blue, $2$ white, $4$ red, $9$ in all), $P(\text{white})$ is
    1. $1/3$
    2. $2/9$
    3. $2/3$
    4. $1/9$
    Check your answer
    1. $1/3$ — The three colours are not equally likely — the nine individual marbles are, giving $P(\text{white}) = 2/9$, not $1/3$.
    2. ✓ $2/9$ — (B) Two of the nine equally likely marbles are white, giving $P(\text{white}) = 2/9$.
    3. $2/3$ — This muddles the white count with blue’s reduced fraction — white alone gives $2/9$.
    4. $1/9$ — There are two white marbles, not one — $P(\text{white}) = 2/9$.
  2. Why are the nine individual marbles equally likely here, but the three colours are not?
    1. each marble is equally likely; the colours are not
    2. because marbles are physical objects, and colours are just names
    3. because there are more marbles than colours
    4. because the bag was shaken before drawing
    Check your answer
    1. ✓ each marble is equally likely; the colours are not — (A) Each marble is equally likely to be drawn, but the colours group unequal numbers of marbles together.
    2. because marbles are physical objects, and colours are just names — Being a physical object versus a name does not explain anything here — what matters is the unequal COUNTS behind each colour.
    3. because there are more marbles than colours — Simply having more marbles than colours does not decide anything — what matters is that the three colours hold unequal counts, $3$, $2$, $4$.
    4. because the bag was shaken before drawing — Shaking the bag changes nothing about the colour counts — the colours are unequal because of how many marbles each holds, not how the bag was handled.
  3. Using the same bag ($3$ blue, $2$ white, $4$ red), what is $P(\text{not red})$?
    1. $4/9$
    2. $1/3$
    3. $1/2$
    4. $5/9$
    Check your answer
    1. $4/9$ — Not-red is not the same size as red — it is $1 - 4/9 = 5/9$, covering the blue and white marbles instead.
    2. $1/3$ — Not-red includes BOTH the blue and white marbles, $3 + 2 = 5$ of the nine — using blue alone gives $1/3$, missing the white marbles.
    3. $1/2$ — The bag does not split evenly — red has $4$ of $9$ marbles, so not-red is $5/9$, not an assumed $1/2$.
    4. ✓ $5/9$ — (D) $P(\text{not red}) = 1 - 4/9 = 5/9$, matching $P(\text{blue}) + P(\text{white}) = 3/9 + 2/9$.
Worked example

Marbles in a box: counting individuals

  1. box holds $3$ blue, $2$ white, $4$ red marbles, $9$ in all, one drawn at random
    The $9$ individual marbles are equally likely, not the $3$ colours.
  2. $P(W) = 2/9$
    $2$ of the $9$ marbles are white.
  3. $P(B) = 3/9 = 1/3$
    $3$ of the $9$ marbles are blue. We divide by $9$, the total marbles, never by the $3$ colours.
  4. $P(R) = 4/9$
    $4$ of the $9$ marbles are red.
  5. $P(W) + P(B) + P(R) = 2/9 + 3/9 + 4/9 = 9/9 = 1$
    *Check that the three probabilities add to $1$.* White, blue and red between them cover every marble in the box, so the three elementary events sum to a whole.
Every marble is one equally likely outcome, and the three colour bands are simply different widths because they hold different numbers of marbles.

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Worked: an ace, or not

Check yourself
  1. One card is drawn from a well-shuffled deck of $52$. $P(\text{ace})$ equals
    1. $1/4$
    2. $4/13$
    3. $1/52$
    4. $1/13$
    Check your answer
    1. $1/4$ — One in four SUITS has any given rank, but the fraction of CARDS that are aces is $4/52 = 1/13$.
    2. $4/13$ — Reducing $4/52$ correctly divides BOTH numbers by $4$, giving $1/13$, not $4/13$.
    3. $1/52$ — There are four aces in the deck, not one — $P(\text{ace}) = 4/52 = 1/13$.
    4. ✓ $1/13$ — (D) Four aces sit among $52$ cards, and $4/52$ reduces to $1/13$.
  2. Why is $4/52$ simplified to $1/13$ rather than left as $4/52$?
    1. because $4/52$ is not a valid probability until reduced
    2. $4/52$ and $1/13$ are the same value, simplified
    3. because the deck actually only has $13$ cards once suits are ignored
    4. because dividing by $4$ accounts for the four aces being drawn one at a time
    Check your answer
    1. because $4/52$ is not a valid probability until reduced — $4/52$ is already a valid probability — reducing it to $1/13$ only changes how it looks, not its value.
    2. ✓ $4/52$ and $1/13$ are the same value, simplified — (B) $4/52$ and $1/13$ are the same value — dividing both numbers by $4$ just writes it more simply.
    3. because the deck actually only has $13$ cards once suits are ignored — The deck still has $52$ cards — $13$ is the number of distinct ranks (ace through king), not a smaller deck size.
    4. because dividing by $4$ accounts for the four aces being drawn one at a time — Dividing by $4$ is purely arithmetic simplification — it says nothing about drawing cards one at a time.
  3. From the same deck, what is $P(\text{not an ace})$?
    1. $1/13$
    2. $9/13$
    3. $11/13$
    4. $12/13$
    Check your answer
    1. $1/13$ — $1/13$ is $P(\text{ace})$ itself — the complement, $P(\text{not an ace})$, is $1 - 1/13 = 12/13$.
    2. $9/13$ — Subtracting $4$ from $13$ is not how the complement rule works — the correct step is $1 - 1/13 = 12/13$.
    3. $11/13$ — Careful subtraction gives $1 - 1/13 = 12/13$, not $11/13$.
    4. ✓ $12/13$ — (D) $P(\text{not an ace}) = 1 - 1/13 = 12/13$.
Worked example

A card draw: an ace, or not

  1. one card drawn from a well-shuffled deck of $52$ cards, all equally likely
    The deck’s $52$ cards are the equally likely outcomes.
  2. $E$ = ‘the card is an ace’: $4$ aces in the deck, so $P(E) = 4/52 = 1/13$
    There is one ace per suit, and four suits.
  3. $P(\text{not E}) = 1 - 1/13 = 12/13$
    This is the complement rule: every non-ace card, found by subtracting from $1$ rather than recounting all $48$ of them directly.
  4. $1/13 + 12/13 = 13/13 = 1$
    *Check that the ace probability and the not-ace probability add back to $1$.* An event and its complement must always sum to a whole deck.
The deck laid out as four rows of thirteen columns highlights the four aces, one in each suit’s row.

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Worked: not green, by the complement rule

Worked example

Not green, by the complement rule

  1. a bag of $3$ green and $7$ blue counters, $10$ in all, one drawn at random
    We list the $10$ counters first: they are the equally likely outcomes, not the two colours.
  2. $P(\text{green}) = 3/10$
    Three of the ten counters are green.
  3. $P(\text{not green}) = 1 - 3/10 = 7/10$
    This is the complement rule: subtract the green probability from $1$, with no second count needed.
  4. $7$ blue counters out of $10$: $P(\text{not green}) = 7/10$
    Check the direct count of blue counters agrees. Seven of the ten counters are blue, which matches $7/10$ found by subtraction.
Find the probability that an event does not happen.
  1. Count the favourable outcomes Count how many outcomes make up the event itself. In the counters example, $3$ of the $10$ counters are green.
  2. Divide to find the probability Divide that count by the total to get the event’s own probability. $P(\text{green}) = 3/10$.
  3. Subtract from one The probability of the event not happening is $1$ minus that answer. $1 - 3/10 = 7/10$.
  4. Check by counting directly Count the outcomes that make up ‘not happening’ on their own, and check the two answers agree. $7$ of the $10$ counters are blue, which matches $7/10$.

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Worked: two dice, a sum of eight

Check yourself
  1. Two dice are thrown together. The probability that the two numbers add up to 8 equals
    1. $1/11$
    2. $4/36$
    3. $5/11$
    4. $5/36$
    Check your answer
    1. $1/11$ — The eleven possible sums are NOT equally likely — the $36$ ordered pairs are. Five of them sum to $8$, giving $5/36$.
    2. $4/36$ — Five pairs sum to $8$, not four — missing $(4,4)$ or any other pair undercounts the favourable outcomes.
    3. $5/11$ — The total is $36$ ordered pairs, not $11$ sums — $5/36$ is correct, not $5/11$.
    4. ✓ $5/36$ — (D) Five pairs, $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(6,2)$, give a sum of $8$, out of $36$ in all.
  2. Why are there $36$ equally likely outcomes to count from, rather than the $11$ possible sums from $2$ to $12$?
    1. because $11$ is simply the wrong count — there are actually $12$ possible sums
    2. the $36$ pairs are equally likely; the sums are not
    3. because two dice can only be thrown one at a time, doubling the outcomes
    4. because a sum of $8$ can only be reached in one particular way
    Check your answer
    1. because $11$ is simply the wrong count — there are actually $12$ possible sums — Eleven sums, $2$ through $12$, is the correct count of possible sums — the real issue is that those eleven sums are not equally likely, unlike the $36$ pairs.
    2. ✓ the $36$ pairs are equally likely; the sums are not — (B) The $36$ ordered pairs are equally likely; the eleven sums are not, since different sums arise from different numbers of pairs.
    3. because two dice can only be thrown one at a time, doubling the outcomes — Throwing the dice together or one after another does not change the outcome count — the $36$ comes from pairing every first-die face with every second-die face.
    4. because a sum of $8$ can only be reached in one particular way — A sum of $8$ is reached by five different pairs, $(2,6)$ through $(6,2)$ — not just one way.
  3. Using the same two dice, how many of the $36$ outcomes give a sum of $10$?
    1. $1$
    2. $4$
    3. $3$
    4. $5$
    Check your answer
    1. $1$ — “Sum of $10$” is reached by three different pairs, $(4,6)$, $(5,5)$, $(6,4)$ — not just one.
    2. $4$ — $(6,6)$ sums to $12$, not $10$ — including it overcounts; the correct count is three pairs.
    3. ✓ $3$ — (C) The pairs $(4,6)$, $(5,5)$, $(6,4)$ give a sum of $10$, so three of the $36$ outcomes qualify.
    4. $5$ — Not every sum has the same number of pairs — sum $8$ has five pairs, but sum $10$ has only three.
  4. The two-coins trap (treating “one head, one tail” as a single outcome) and the mistake of treating every dice-sum as equally likely are the same kind of error. What do the two mistakes share?
    1. both come from using a die and a coin in the same problem
    2. both collapse several outcomes into one
    3. both are fixed by tossing more coins or throwing more dice
    4. both only happen when the two coins or two dice look identical
    Check your answer
    1. both come from using a die and a coin in the same problem — These are two separate mistakes, one about coins and one about dice — no problem here mixes a die and a coin together.
    2. ✓ both collapse several outcomes into one — (B) Both mistakes collapse several equally likely outcomes — $H T$/$T H$, or several dice pairs — into a single counted case.
    3. both are fixed by tossing more coins or throwing more dice — Repeating the trial more times does not fix either mistake — the fix is counting outcomes at the right grain from the start.
    4. both only happen when the two coins or two dice look identical — Whether the objects look identical is irrelevant — the error is in how the outcomes are grouped, not in their appearance.
Worked example

Two dice: a sum of eight

  1. two dice, $36$ equally likely outcomes since $6 \cdot 6 = 36$
    Each die has six faces, and both dice are thrown together.
  2. $E$ = ‘the sum is $8$’ has favourable pairs $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$, $(6,2)$
    We list every ordered pair that sums to $8$, none omitted and none repeated.
  3. $P(E) = 5/36$
    Five favourable pairs out of the $36$ equally likely ordered pairs.
  4. $0 \leq 5/36 \leq 1$
    *Check that the answer falls inside $0$ to $1$.* Every probability answer must fall inside this range, whenever a count is finished.
Fixing the total at eight picks out a slanting line of five cells, since raising one die while lowering the other keeps the sum unchanged.
A girl in a green cap has just thrown two dice on a grey desk; the white die shows four on top and the pale blue die shows five.
  • 4 on this die
  • 5 on this one
Tara throws two dice and gets 4 and 5: one of the 36 throws, each as likely as any other.
Find the probability of an event on two dice.
  1. Count the total outcomes Two dice give $6 \cdot 6 = 36$ ordered pairs, since each die has six faces.
  2. Decide which pairs the event wants Pick the rule the event sets, such as the two faces summing to $8$.
  3. List every pair in both orders Write out every ordered pair that satisfies it, such as $(2,6)$ and $(6,2)$. These are two different pairs, not one.
  4. Divide by thirty-six Divide the count of favourable pairs by $36$, the total. Five pairs sum to $8$, so $P(E) = 5/36$.
  5. Check by a second route Read the first die from $2$ to $6$ and see that each value has exactly one partner making $8$. Five values give five pairs.

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Recap

■RECAP
Check yourself
  1. This chapter’s opening claim is that every probability problem reduces to one formula. What is that formula, in words?
    1. favourable outcomes over total outcomes
    2. count the total outcomes, then subtract the favourable ones
    3. list every possible outcome and pick the most likely one
    4. average the number of favourable and unfavourable outcomes
    Check your answer
    1. ✓ favourable outcomes over total outcomes — (A) The one formula is: count the favourable outcomes, count all possible outcomes, and divide.
    2. count the total outcomes, then subtract the favourable ones — Subtracting gives the count of unfavourable outcomes, not a probability — the formula divides the two counts instead.
    3. list every possible outcome and pick the most likely one — The formula gives a numeric probability for a chosen event — it does not ask you to pick out the single most likely outcome.
    4. average the number of favourable and unfavourable outcomes — Averaging favourable and unfavourable counts is not the formula — it divides favourable outcomes by the TOTAL, not an average.
  2. Across this chapter’s coin, die, card, marble, and dice-pair examples, what is the one skill that actually varies from problem to problem?
    1. which formula to apply, since each situation uses a different one
    2. whether to add or multiply the final probabilities
    3. finding the right counts each time
    4. how many decimal places to round the final probability to
    Check your answer
    1. which formula to apply, since each situation uses a different one — The same formula is used throughout this chapter — coins, dice, cards, marbles all use favourable-over-total; only the counts change.
    2. whether to add or multiply the final probabilities — None of these examples ever combine final probabilities by adding or multiplying — each is a single favourable-over-total computation.
    3. ✓ finding the right counts each time — (C) The formula stays fixed; what changes each time is finding the correct favourable and total counts.
    4. how many decimal places to round the final probability to — This chapter’s probabilities are left as exact fractions — rounding is not a step any of these examples use.
  3. A problem gives outcomes that are not confirmed to be equally likely. Before applying this chapter’s formula, what must be done first?
    1. check that the outcomes really are equally likely
    2. apply the formula anyway, since it always gives some answer
    3. switch to counting sums instead of individual outcomes
    4. repeat the experiment several times to see which outcome happens most often
    Check your answer
    1. ✓ check that the outcomes really are equally likely — (A) The formula only applies once the individual outcomes are confirmed equally likely, breaking down any category first if needed.
    2. apply the formula anyway, since it always gives some answer — The formula always produces a number, but that number is only a correct probability once the counted outcomes are confirmed equally likely.
    3. switch to counting sums instead of individual outcomes — Counting sums instead of individual outcomes is the mistake this chapter warns against, not a fix — the fix is checking at the individual-outcome level.
    4. repeat the experiment several times to see which outcome happens most often — This chapter’s probability comes from counting outcomes in advance, not from observing repeated trials — that is a different idea from another course.
Check yourself: the whole chapter
  1. This chapter’s rule is for equally likely outcomes. It finds the probability of an event by
    1. multiplying the favourable outcome count by the total outcome count
    2. dividing the total outcome count by the favourable outcome count
    3. dividing the favourable outcome count by the total outcome count
    4. subtracting the favourable outcome count from the total outcome count
    Check your answer
    1. multiplying the favourable outcome count by the total outcome count — Multiplying the favourable count by the total gives a number far outside the valid range for a probability — the rule divides, it never multiplies.
    2. dividing the total outcome count by the favourable outcome count — The favourable count belongs on top of the fraction and the total belongs on the bottom — this option puts them the wrong way round.
    3. ✓ dividing the favourable outcome count by the total outcome count — (C) $P(E)$ divides the count of favourable outcomes by the count of all possible outcomes, once every outcome is equally likely.
    4. subtracting the favourable outcome count from the total outcome count — Subtracting one count from the other gives a count of outcomes, not a probability — the rule divides the two counts instead.
  2. A bag contains $5$ red balls and $3$ blue balls, all equally likely to be drawn. What is $P(\text{red})$, the probability of drawing a red ball?
    1. $3/8$, the fraction of blue balls instead of red
    2. $8/5$, the total number of balls divided by the number of red balls
    3. $5$, the number of red balls, without dividing by the total
    4. $5/8$, the number of red balls divided by the total number of balls
    Check your answer
    1. $3/8$, the fraction of blue balls instead of red — The question asks for $P(\text{red})$, so the red-ball count, $5$, belongs on top — not the blue-ball count.
    2. $8/5$, the total number of balls divided by the number of red balls — Dividing the total by the favourable count inverts the fraction; the favourable count belongs on top, the total on the bottom.
    3. $5$, the number of red balls, without dividing by the total — Counting the red balls alone is only half the calculation — the count still needs to be divided by the total number of balls, $8$.
    4. ✓ $5/8$, the number of red balls divided by the total number of balls — (D) There are $8$ balls in all, and $5$ of them are red, so $P(\text{red}) = 5/8$.
  3. A bag has $4$ red marbles, $1$ blue marble and $1$ green marble, all equally likely to be picked. What is $P(\text{blue})$?
    1. $1/3$, treating each of the three colours as equally likely, instead of each marble
    2. $1/6$, one blue marble out of six total marbles
    3. $1/4$, comparing the blue marble only to the red marbles, ignoring the green marble
    4. $1$, counting the blue marble without comparing it to the rest of the bag
    Check your answer
    1. $1/3$, treating each of the three colours as equally likely, instead of each marble — The three colours are not equally likely — the bag holds unequal numbers of each. It is the six individual marbles, one at a time, that are equally likely.
    2. ✓ $1/6$, one blue marble out of six total marbles — (B) The six individual marbles are equally likely, not the three colours — one of the six is blue, so $P(\text{blue}) = 1/6$.
    3. $1/4$, comparing the blue marble only to the red marbles, ignoring the green marble — The total must include every marble in the bag, all six of them, not just the red ones — leaving green out understates the total.
    4. $1$, counting the blue marble without comparing it to the rest of the bag — Counting the blue marble alone is only half the calculation — it still needs to be divided by the total number of marbles, $6$.
  4. The probability of an event $E$ is $P(E) = 2/5$. What is the probability of ‘not $E$’, the complement of $E$?
    1. $2/5$, the same as $P(E)$, since the complement just describes the same event differently
    2. $5/2$, the reciprocal of $P(E)$
    3. $3/5$, one minus $P(E)$, since $P(E)$ and $P(\text{not} E)$ always add to one
    4. $1/5$, half of $P(E)$
    Check your answer
    1. $2/5$, the same as $P(E)$, since the complement just describes the same event differently — ‘Not $E$’ happens exactly when $E$ does not — a different event from $E$ itself, even though the two probabilities are related.
    2. $5/2$, the reciprocal of $P(E)$ — Finding the complement means subtracting $P(E)$ from $1$, not inverting the fraction.
    3. ✓ $3/5$, one minus $P(E)$, since $P(E)$ and $P(\text{not} E)$ always add to one — (C) $P(E)$ and $P(\text{not}E)$ always add to $1$, so $P(\text{not}E) = 1 - 2/5 = 3/5$.
    4. $1/5$, half of $P(E)$ — There is no halving rule for complements — the two probabilities add to $1$, so $P(\text{not}E) = 1 - P(E)$ directly.
  5. A calculation for the probability of an event gives $P(E) = 1.2$. What does this tell you?
    1. the event is more than certain to happen
    2. the event is impossible, since a value above $1$ signals a contradiction with certainty
    3. the answer is correct, since probability can range from $0$ to any positive number
    4. a mistake has been made somewhere, since probability can never exceed $1$
    Check your answer
    1. the event is more than certain to happen — Certain is already the top of the scale, $P(E) = 1$ — there is no ‘more than certain’ category above it.
    2. the event is impossible, since a value above $1$ signals a contradiction with certainty — An impossible event has $P(E) = 0$, the opposite end of the scale from $1.2$ — a too-large result is not the same as impossibility.
    3. the answer is correct, since probability can range from $0$ to any positive number — Probability is always between $0$ and $1$ by definition — a value of $1.2$ cannot be a genuine probability, whatever produced it.
    4. ✓ a mistake has been made somewhere, since probability can never exceed $1$ — (D) $0 \leq P(E) \leq 1$ always — a result of $1.2$ falls outside that range, so an error has been made somewhere in the calculation.

We count every probability here the same way: outcomes divided by outcomes, applied to the sample space. Try one last check yourself: recount any worked example above from scratch, and see that you land on the same answer.

The same ten tickets answer all three questions, and only the count of favourable ones changes from row to row.

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Where you will meet this

You judge a chance more often than you think, from a raffle ticket to a bus seat. Here are seven places the probability formula gives you a real number.

Two of twelve identical jars are turmeric, giving a probability of 2 over 12, which is 1/6.

Your turn. A parking lot has $60$ spaces, and $15$ are reserved for staff. If a space is chosen at random, what is the chance it is a staff space? Answer: The chance is $15/60 = 1/4$.

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Practice set: Exercise 14.1

Exercise 14.1
  1. practice A bag contains $4$ red and $8$ green counters, identical apart from colour. One counter is drawn at random. Find the probability that it is red, and the probability that it is not red. (Worked in full below — read it, then do the next two the same way.)
  2. practice A tray holds $8$ mangoes, of which $2$ are unripe. One mango is picked at random. Find the probability that it is unripe, and the probability that it is not unripe. (The same two steps as above — the total first, then subtract from $1$.)
  3. practice A bag contains only white and black beads. The probability of drawing a white bead at random is $3/7$. Find the probability of drawing a black bead. (Nothing to count here — the first probability is already given, so the rule does the rest.)
  4. practice A fair coin is tossed once. Find the probability of getting a tail.
  5. practice A die is thrown once. Find the probability of getting a number less than $3$.
  6. practice A card is drawn at random from a well-shuffled deck of $52$ playing cards. Find the probability that the card drawn is a king.
  7. practice A card is drawn at random from a well-shuffled deck of $52$ playing cards. Find the probability that the card drawn is not a face card.
  8. practice A bag contains $5$ red, $3$ green and $2$ yellow discs, identical apart from colour. One disc is drawn at random. Find the probability that it is green.
  9. practice A box contains $6$ good pens and $4$ defective pens. One pen is drawn at random. Find the probability that the pen drawn is defective.
  10. practice A carton has $20$ bulbs, of which $3$ are defective. One bulb is drawn at random. Find the probability that it is not defective.
  11. practice Two coins are tossed together. Find the probability of getting at least one head.
  12. practice Two dice are thrown together. Find the probability that the sum of the numbers on the two dice is $10$.
  13. practice A bag contains only red and blue marbles. The probability of drawing a red marble is $2/5$. Find the probability of drawing a blue marble.
  14. practice A bag contains only red balls. One ball is drawn at random. Find the probability that the ball drawn is blue.
  15. practice A spinner is divided into $8$ equal sectors numbered $1$ to $8$. Find the probability that the spinner stops on a number less than $9$.
  16. practice State whether the statement is true or false, and justify: ‘The probability of an event is always a whole number.’
  17. practice A student says: ‘A die has six faces, numbered $1$ to $6$. Since $6$ is even, the probability of an even number is $1/6$.’ Find the error, and give the correct probability.
  18. practice A die is thrown once. Verify that the probabilities of its six elementary events add up to $1$.
Answers
  1. $P(\text{red}) = 4/12 = 1/3$, and $P(\text{not red}) = 1 - 1/3 = 2/3$ — counting the green counters directly gives $8/12 = 2/3$ as well.
  2. $P(\text{unripe}) = 2/8 = 1/4$, and $P(\text{not unripe}) = 1 - 1/4 = 3/4$.
  3. $P(\text{black}) = 1 - 3/7 = 4/7$.
  4. $1$ favourable outcome out of $2$: $P(\text{tail}) = 1/2$.
  5. Numbers less than $3$ are $1, 2$: $P(E) = 2/6 = 1/3$.
  6. $4$ kings in the deck: $P(\text{king}) = 4/52 = 1/13$.
  7. Face cards number $12$ ($4$ jacks, $4$ queens, $4$ kings): $P(\text{face}) = 12/52 = 3/13$, so $P(\text{not face}) = 1 - 3/13 = 10/13$.
  8. $5 + 3 + 2 = 10$ discs in all: $P(\text{green}) = 3/10$.
  9. $6 + 4 = 10$ pens in all: $P(\text{defective}) = 4/10 = 2/5$.
  10. $P(\text{defective}) = 3/20$, so $P(\text{not defective}) = 1 - 3/20 = 17/20$.
  11. Outcomes $H H$, $H T$, $T H$, $T T$; at least one head is $H H$, $H T$, $T H$: $P(E) = 3/4$.
  12. Pairs summing to $10$: $(4,6)$, $(5,5)$, $(6,4)$: $P(E) = 3/36 = 1/12$.
  13. $P(\text{red}) + P(\text{blue}) = 1$, so $P(\text{blue}) = 1 - 2/5 = 3/5$.
  14. No blue ball is in the bag, so drawing one is an impossible event: $P(\text{blue}) = 0$.
  15. Every one of the $8$ numbered sectors shows a number less than $9$, so this is a sure event: $P(E) = 8/8 = 1$.
  16. False. A probability satisfies $0 \leq P(E) \leq 1$ and is a whole number only at the two extremes, $0$ and $1$; most probabilities, like $1/2$ or $3/10$, are not whole numbers.
  17. The error: $6$ being even is a fact about one face’s label, not a count of favourable outcomes. The even-numbered faces are $2, 4, 6$, three faces out of six, so $P(\text{even}) = 3/6 = 1/2$.
  18. Each face has probability $1/6$: $1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1$.
A row and column of six overlap in one cell, so eleven cells mark at least one six, with the shared cell outlined.
Cells with a total of ten or more cluster into a solid corner of the grid, rather than fall along a single line.
Exercise 14.1 — further practice
  1. practice A fair die, with faces numbered $1$ to $6$, is thrown once. Find the probability of getting a multiple of $3$.
  2. practice One card is drawn at random from a well-shuffled deck of $52$ playing cards. Find the probability that the card is a black king.
  3. practice A box contains $50$ discs, numbered $1$ to $50$, one number per disc. One disc is drawn at random. Find the probability that the number on it is divisible by $7$.
  4. practice A spinner is divided into $5$ equal sectors, numbered $1$ to $5$. Find the probability that the spinner stops on an odd number.
  5. practice A cricket team has $11$ players, wearing jersey numbers $1$ to $11$. One player is chosen at random to be the captain. Find the probability that the captain’s jersey number is greater than $8$.
  6. practice Which of the following values cannot be the probability of an event?
    1. $0.2$
    2. $-0.4$
    3. $3/5$
    4. $1$
  7. practice A sweet shop has a box of $30$ ladoos: $12$ besan, $10$ motichoor and $8$ kaju. One ladoo is picked at random from the box. Find the probability that it is not a motichoor ladoo.
  8. practice A person is equally likely to be born on any day of the week. Find the probability that the day of birth is a day whose name starts with the letter $S$.
  9. practice A jar holds $9$ blue and $15$ pink glass bangles, all the same size. One bangle is picked at random from the jar. Find the probability that it is not pink.
  10. practice If $P(E) = 0.37$ for an event $E$, what is $P(\text{not E})$?
    1. $0.63$
    2. $1.37$
    3. $-0.37$
    4. $0.37$
  11. practice A fair die, with faces numbered $1$ to $6$, is thrown once. Find the probability of getting a number that is both even and prime.
  12. practice One card is drawn at random from a well-shuffled deck of $52$ playing cards. Find the probability that the card drawn shows a number from $2$ to $10$, not an ace or a face card.
  13. practice Two fair coins are tossed together. Find the probability of getting two tails.
  14. practice A number is chosen at random from the integers $1$ to $10$. What is the probability that it is a multiple of $4$?
    1. $1/10$
    2. $1/5$
    3. $3/10$
    4. $2/5$
  15. practice Two fair dice, each numbered $1$ to $6$, are thrown together. Find the probability that the sum of the numbers on the two dice is $9$.
  16. practice Two fair dice, each numbered $1$ to $6$, are thrown together. Find the probability that the sum of the numbers on the two dice is at least $11$.
  17. practice Two fair dice, each numbered $1$ to $6$, are thrown together. Find the probability that the sum of the numbers on the two dice is a prime number.
  18. practice A box contains $30$ cards, numbered $1$ to $30$, one number per card. One card is drawn at random. Find the probability that the number on it is neither a multiple of $3$ nor a multiple of $5$.
Answers
  1. $P(E) = 1/3$
  2. $P(E) = 1/26$
  3. $P(E) = 7/50$
  4. $P(E) = 3/5$
  5. $P(E) = 3/11$
  6. B — $-0.4$.
  7. $P(E) = 2/3$
  8. $P(E) = 2/7$
  9. $P(E) = 3/8$
  10. A — $0.63$.
  11. $P(E) = 1/6$
  12. $P(E) = 9/13$
  13. $P(E) = 1/4$
  14. B — $1/5$.
  15. $P(E) = 1/9$
  16. $P(E) = 1/12$
  17. $P(E) = 5/12$
  18. $P(E) = 8/15$

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