Probability

From certainty to likelihood

Your cousin shuffles a pack of 52 cards. She draws one and lays it face down on the table. “Is it a king?” she asks. Four of the 52 cards are kings, so the chance is $4/52 = 1/13$. Then she lifts one corner, looks, and says only: “It is a picture card.” A picture card is a jack, a queen or a king (also called a face card). What is the chance of a king now?

before

a picture card

It is not 1 in 13 any more. The card has not changed. What has changed is what you know. Only the 12 picture cards are still possible, and 4 of them are kings. So the chance is now $4/12 = 1/3$.

In Class 10 and Class 11 you asked how likely an event is when nothing else is known. You counted the outcomes and divided. But most real questions come after you learn something. A test result comes back. Your cousin says the card is a picture card. A team scores first.

Here is the question to carry to the last page. How does a chance change when you learn something? And how do you work back from what you saw to what most likely caused it?

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Reading the notation

You have a card still lying face down. Its chance of being a king changed from 1 in 13 to 1 in 3 when you heard one fact. This section gives you the symbols to write that change. Each symbol is a short way to say a sentence you can already say. One picture carries all of them: the 36 outcomes of two dice.

One grid, two events

Throw two dice, a first and a second. Each outcome is a pair, such as $(4, 4)$. There are $6 \times 6 = 36$ pairs, all equally likely, and together they make the sample space $S$. Read $S$ as “every outcome”. Now name two events. $A$ is “the total is 8”. $B$ is “the first die shows 4”.

*Read $P(A)$ as* “the probability of $A$”. To find it, count the outcomes of $A$ and divide by all 36. $A$ holds $(2, 6)$, $(3, 5)$, $(4, 4)$, $(5, 3)$ and $(6, 2)$, so $P(A) = 5/36$. $B$ is the whole of row 4, six outcomes, so $P(B) = 6/36 = 1/6$.

*Read $A \cap B$ as* “$A$ and $B$”: both happen at once. Only one cell is in both, $(4, 4)$, so $P(A \cap B) = 1/36$.

*Read $A \cup B$ as* “$A$ or $B$, or both”. Count the five cells of $A$ and the six of $B$, but count the shared cell $(4, 4)$ once. That gives $5 + 6 - 1 = 10$ cells, so $P(A \cup B) = 10/36$. This is the addition rule of Class 11: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

*Read $A'$ as* “not $A$”. It holds every outcome that is not in $A$: the 31 cells outside the five. So $P(A') = 31/36$. You can also get it from $P(A)$, since $A$ and $A'$ together fill $S$: $P(A') = 1 - 5/36 = 31/36$.

The bar is a wall

Now suppose you are told that the first die shows 4. You do not know the second die yet. Write the new question as $P(A|B)$, and read it as “the probability of $A$, given $B$”. The bar is a wall: whatever stands to its right is known, and the sample space reduces to it.

Count only the outcomes in B. Only row 4 is left, six equally likely outcomes. Of those, one adds to 8: the cell $(4, 4)$. So $P(A|B) = 1/6$. Before you knew $B$, the chance of a total of 8 was $5/36$. Knowing $B$ has changed it, as the picture card changed the chance of a king.

It is a share. $P(A|B)$ is never “$A$ divided by $B$”. It is never $P(A \cap B)$ either: it is $A$’s share of $B$. Here $P(A \cap B) = 1/36$, the shared cell’s share of all 36 cells. But $P(A|B) = 1/6$, the same cell’s share of row 4 alone. The cell is the same. What you divide by is not.

Now turn it round. Suppose you are told only that the total is 8. Then the sample space reduces to the five cells of $A$. One of them, $(4, 4)$, has a first die of 4, so $P(B|A) = 1/5$. $P(A|B)$ and $P(B|A)$ are different questions with different answers: 1 in 6 against 1 in 5. The two grids side by side are the proof.

Here is every symbol of this section in one table. Cover the last two columns and fill them in from the grid.

The card in symbols

Go back to the card. Let $K$ be “the card is a king”, and let $F$ be “the card is a picture card”. Then $P(K) = 4/52$ and $P(F) = 12/52$. Every king is a picture card, so $K \cap F$ is just the four kings, and $P(K \cap F) = 4/52$. Given $F$, only the 12 picture cards are left, and 4 of them are kings. So $P(K|F) = 4/12 = 1/3$. That is the change from 1 in 13 to 1 in 3, written in symbols.

Counting works here because every outcome is equally likely and you can list them all. The next section makes “given” exact for any two events, with one formula.

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Probability, once you know something

A friend throws a die and hides it under a cup. What is the chance that it shows a 6? The six faces are equally likely, so it is 1 in 6.

Now your friend peeks and says, “It is even.” Three faces are left: 2, 4 and 6. They are still equally likely. So the chance of a 6 is now 1 in 3. The new fact reduced the sample space from six outcomes to three: this narrower set is the REDUCED SAMPLE SPACE, the board’s term for it.

Let us write that with events, in the symbols of the last section. Let $B$ be the event you are told has happened. Let $A$ be the event you ask about. Once $B$ is known, only the outcomes in $B$ count. Of those, the ones that also lie in $A$ make up $A \cap B$.

A ratio of chances

The CONDITIONAL PROBABILITY of $A$ given $B$ is $P(A|B) = P(A \cap B)/P(B)$, provided $P(B) \neq 0$.

You read $P(A|B)$ as “the probability of $A$, given $B$”. The bar is not a division. If $P(B) = 0$, there is nothing to divide by. Then $P(A|B)$ is not defined.

Check it on the die. Here $A$ is “a 6” and $B$ is “an even number”. So $A \cap B = \{6\}$, with $P(A \cap B) = 1/6$ and $P(B) = 3/6$. Then $P(A|B) = (1/6)/(3/6) = 1/3$, as you counted.

The figure draws the same idea. On the left, $P(A)$ is the share of $A$ in the whole space. On the right, you are given $B$, so the part of $A$ outside $B$ is greyed out. Now $P(A|B)$ is the share of $B$ that the overlap takes up.

Reduce the sample space

Your turn: a die shows a number greater than 3. What is the chance that it is even?

A conditional probability is still a probability. It uses $B$ as its new whole. Take any event $B$ with $P(B) \neq 0$. Three facts follow from the definition.

First, $P(S|B) = P(B|B) = 1$. Once you know that $B$ happened, $B$ is certain. So the whole space $S$ is certain too.

Second, $0 \leq P(A|B) \leq 1$ for every event $A$. The part $A \cap B$ lies inside $B$. So its chance can never be more than the chance of $B$.

Third, the complement rule still works. Here $A'$ is the event “not $A$”. Then $P(A'|B) = 1 - P(A|B)$. On the die, given “even”, a 6 has chance $1/3$. So “not a 6” has chance $2/3$. Those are the faces 2 and 4.

The addition rule carries over as well. For any events $A$ and $C$, $P((A \cup C)|B) = P(A|B) + P(C|B) - P((A \cap C)|B)$. If $A$ and $C$ cannot both happen, the last term is 0.

The figure shows the first and third facts. On the left, $B$ is the whole, and its share is 1. On the right, $A$ cuts $B$ in two. The two shares, $P(A|B)$ and $P(A'|B)$, add to 1.

In each fact, $B$ stays fixed. If you change the event after the bar, you change the question.

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Turning the definition around

A box of sweets has 4 in red wrappers and 2 in blue. You take one without looking and eat it. Then you take a second. What is the chance that both are red?

The first pick is easy. Four of the six are red, so the chance is 4 in 6, or 2 in 3. The second pick depends on the first. If the first was red, the box now holds 3 red and 2 blue. So the chance of a second red is 3 in 5.

How do the two chances combine? A red comes first in 2 of every 3 tries. Of those, a second red follows 3 times in 5. So both are red in 3/5 of 2/3 of all tries.

Let us write that with events. Let $R_1$ be “the first is red” and $R_2$ be “the second is red”. Then $P(R_1 \cap R_2) = P(R_1) \cdot P(R_2|R_1) = 2/3 \cdot 3/5 = 2/5$.

This is the definition of conditional probability, turned around. Multiply both sides of $P(A|B) = P(A \cap B)/P(B)$ by $P(B)$.

The rule for two events

The MULTIPLICATION THEOREM says that for events $A$ and $B$ with $P(A) \neq 0$ and $P(B) \neq 0$, $P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)$. Either event can go first in the formula. The answer is the same.

For three events, each new event is taken as given all the ones before it: $P(A \cap B \cap C) = P(A) \cdot P(B|A) \cdot P(C|(A \cap B))$, provided $P(A \cap B) > 0$.

The figure draws the same thing as a bag of 4 red balls and 2 blue. The first strip shows red as 2 parts in 3. The second shows red as 3 parts in 5, since no ball goes back. Together, both red is 6 parts in 15.

Now say the first sweet went back into the box. The second pick would see 4 red in 6 again. Then the first pick would change nothing at all. Such events are called independent, the next idea.

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Independent events

Throw a coin and a die at the same time. Does the coin change the die’s chances? No: tossing the coin has no effect on the die.

So say the coin shows heads. The chance of a 6 is still 1 in 6. Learning about the coin told you nothing about the die.

Let us check that with the formula. Let $A$ be “heads” and $B$ be “a 6”. There are 12 equally likely outcomes, from H1 to T6. Only H6 is in both, so $P(A \cap B) = 1/12$. Then $P(B|A) = (1/12)/(1/2) = 1/6$. That is $P(B)$ again.

The product test

Two events $A$ and $B$ are INDEPENDENT EVENTS if $P(A \cap B) = P(A) \cdot P(B)$. When $P(A) \neq 0$, this is the same as $P(B|A) = P(B)$. When $P(B) \neq 0$, it is the same as $P(A|B) = P(A)$. Learning one event leaves the chance of the other as it was.

For the coin and the die, $P(A) \cdot P(B) = 1/2 \cdot 1/6 = 1/12$. That equals $P(A \cap B)$, so they are independent.

You decide independence from the numbers, not from the story. Events can sound linked and still pass the test. The worked example below has a pair like that.

The figure draws independence in a square. $A$ is a column of width 0.6, and $P(B) = 0.35$. On the left, $B$ is cut straight across both columns. Then $P(A \cap B) = 0.6 \cdot 0.35 = 0.21$, so the events are independent. On the right, the cut jogs. Now $P(A \cap B)$ is 0.30, not 0.21, so they are dependent.

One more fact. If $A$ and $B$ are independent, so are $A$ and $B'$. Here is why.

1. $A$ splits into $A \cap B$ and $A \cap B'$, and the two cannot both happen.
2. $P(A \cap B') = P(A) - P(A \cap B)$, because the two pieces add up to $P(A)$.
3. $P(A \cap B) = P(A) \cdot P(B)$, because $A$ and $B$ are independent.
4. $P(A \cap B') = P(A) \cdot (1 - P(B)) = P(A) \cdot P(B')$, factoring out $P(A)$ and using $P(B') = 1 - P(B)$.

*Swap $A$ and $B$ in the same argument,* and $A'$ and $B$ are independent too:

1. $B$ splits into $B \cap A$ and $B \cap A'$, and the two cannot both happen.
2. $P(B \cap A') = P(B) - P(B \cap A) = P(B) - P(A) \cdot P(B) = P(B) \cdot (1 - P(A)) = P(B) \cdot P(A')$.

*Apply the same result to $A'$ and $B$,* splitting by $B'$ this time, and $A'$ and $B'$ are independent too:

1. $A'$ splits into $A' \cap B$ and $A' \cap B'$, and the two cannot both happen.
2. $P(A' \cap B') = P(A') - P(A' \cap B) = P(A') - P(A') \cdot P(B) = P(A') \cdot (1 - P(B)) = P(A') \cdot P(B')$.

Mutually exclusive

Two events are MUTUALLY EXCLUSIVE if they cannot both happen. Then $A \cap B = \emptyset$, so $P(A \cap B) = 0$.

First thought

?

Joint chance

P(odd and even)

0

Product

1/2 × 1/2

1/4

not 0

Independent is a different idea. It says that knowing one event does not change the chance of the other. The two words can sound alike. They mean different things.

Take two events with $P(A) \neq 0$ and $P(B) \neq 0$. Say they are mutually exclusive. Then $P(A \cap B) = 0$. But $P(A) \cdot P(B)$ is more than 0. The product test fails, so they are not independent.

If $A$ happens, you know for sure that $B$ did not.

The figure puts the two cases side by side. On the left, the circles $A$ and $B$ do not meet. So $P(A \cap B) = 0$, while $P(A) \cdot P(B)$ is more than 0. They are never independent. On the right, the circles overlap. They are independent only if the overlap equals the product.

Here is a pair that can happen together. Throw a die. Let $A$ be “an odd number” and $B$ be “a number less than 3”. Then $A \cap B = \{1\}$, so $P(A \cap B) = 1/6$. And $P(A) \cdot P(B) = 1/2 \cdot 1/3 = 1/6$. So these two are independent.

Test two faces

Your turn: throw a die. Let $A$ be “a 1” and $B$ be “a 2”. Find $P(A \cap B)$ and $P(A) \cdot P(B)$. Are $A$ and $B$ independent?

Three events $A$, $B$ and $C$ are mutually independent if all four of these hold. - $P(A \cap B) = P(A) \cdot P(B)$. - $P(A \cap C) = P(A) \cdot P(C)$. - $P(B \cap C) = P(B) \cdot P(C)$. - $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$. If even one of them fails, you cannot call the three independent.

Why check the last one? The three pairs can pass while the triple fails. Toss a coin twice. Let $A$ be “a head on the first toss” and $B$ be “a head on the second”. Let $C$ be “the two tosses match”. The four outcomes HH, HT, TH and TT are equally likely.

Each event has chance $1/2$. Each pair meets only in HH, with chance $1/4 = 1/2 \cdot 1/2$. So every pair passes the test. But all three also meet only in HH. So $P(A \cap B \cap C) = 1/4$, while the product is $1/8$. The three events are not independent.

For independent events $A_1, A_2, \ldots, A_n$, the chances multiply: $P(A_1 \cap A_2 \cap \ldots \cap A_n) = P(A_1) \cdot P(A_2) \cdot \ldots \cdot P(A_n)$. Here independent means the product rule holds for every group of the events, not only for the pairs.

The figure shows how fast such a product shrinks. Each step works with chance 0.8. Two steps give 0.64, and three give 0.512. Four steps give 0.4096, which is below an even chance of 0.5. Five give 0.32768.

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Splitting the sample space into cases

Two bags sit on a table. Bag A holds 2 red balls and 1 white. Bag B holds 1 red ball and 2 white. You toss a coin. Heads, you draw a ball from bag A. Tails, you draw from bag B. What is the chance that the ball is red?

Think about how a red ball can come. The coin can pick bag A, and the ball can be red. Or the coin can pick bag B, and the ball can be red. These two ways cannot both happen. And there is no third way, since the coin picks one bag every time.

Splitting into cases like this is a PARTITION. Events $E_1, E_2, \ldots, E_n$ form a partition of the sample space $S$ if three things hold. - They are pairwise disjoint: $E_i \cap E_j = \emptyset$ for $i \neq j$. - Together they make up $S$: $E_1 \cup E_2 \cup \ldots \cup E_n = S$. - Each one has $P(E_i) > 0$. Then every outcome falls in exactly one of them.

For the bags, $E_1$ is “the coin picks bag A” and $E_2$ is “the coin picks bag B”. Each has chance $1/2$. They cannot both happen, and one of them always does. So they form a partition.

The figure shows a partition as one strip. It is cut into $E_1$, $E_2$ and $E_3$, of widths 5/10, 3/10 and 2/10. They add to 1. Below it are two splits that fail. In one, two pieces overlap. In the other, a gap is left.

A split to check

Your turn: throw a die. Let $E_1$ be “an even number” and $E_2$ be “a multiple of 3”. Do $E_1$ and $E_2$ form a partition?

Back to the two bags. How likely is a red ball?

Split by the bag. The coin picks bag A half the time, and then red comes 2 times in 3. It picks bag B half the time, and then red comes 1 time in 3. Add the two ways. Let $R$ be “the ball is red”. Then $P(R) = 1/2 \cdot 2/3 + 1/2 \cdot 1/3 = 1/3 + 1/6 = 1/2$.

Each term is the multiplication theorem for one case. The cases cannot overlap, so their shares just add.

Adding over every case

Here is the THEOREM OF TOTAL PROBABILITY. Let $E_1, E_2, \ldots, E_n$ be a partition of $S$, with each $P(E_i) > 0$. Then for any event $A$, $P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) + \ldots + P(E_n) \cdot P(A|E_n)$.

Why is it true? The event $A$ splits into the pieces $A \cap E_1$ to $A \cap E_n$. No two pieces overlap, and together they make up $A$. So $P(A)$ is the total of their chances. By the multiplication theorem, each piece has chance $P(E_i) \cdot P(A|E_i)$.

The figure works one more case. $E_1$ has chance $1/4$ and $E_2$ has $3/4$. Inside $E_1$, $A$ happens 2 times in 3. Inside $E_2$, it happens 1 time in 3. So $P(A) = 2/12 + 3/12 = 5/12$.

You will use this total again in a moment. It is the denominator of Bayes’ theorem.

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Bayes’ theorem

In 2020 and 2021, many people took a test for Covid. Say your test comes back positive. Does that mean you are ill? Not always. Let us work it out with small numbers.

Say 10 people in 1000 have an illness. A test for it is right 90% of the time, for the ill and for the well. Of the 10 ill people, 9 test positive. Of the 990 well people, 10% test positive by mistake. That is 99 people. So 108 people test positive, and only 9 of them are ill. Given a positive test, the chance of the illness is 9 in 108, or 1 in 12.

Most of the positives come from the well. There are so many more well people that their few mistakes outnumber the ill.

You just worked backwards. You knew the chance of a positive test, given the illness. You found the chance of the illness, given a positive test. These are different numbers: 9 in 10 and 1 in 12.

Now in symbols

BAYES’ THEOREM does this for any partition. Let $E_1, E_2, \ldots, E_n$ be a partition of $S$, with each $P(E_i) > 0$. Let $A$ be any event with $P(A) \neq 0$. Then $P(E_i|A) = (P(E_i) \cdot P(A|E_i))/(\sum_{j=1}^n P(E_j) \cdot P(A|E_j))$.

before the test

after a positive

The top of Bayes’ fraction, $P(E_i) \cdot P(A|E_i)$, is the chance that case $E_i$ and $A$ both happen. The bottom is $P(A)$, found by total probability over every case. Each $P(E_i)$ is a PRIOR: the chance before you learn $A$. Each $P(E_i|A)$ is a POSTERIOR: the chance after you learn it.

Check it on the test. Let $E_1$ be “ill” and $E_2$ be “well”, with $P(E_1) = 1/100$ and $P(E_2) = 99/100$. Let $A$ be “tests positive”. Then $P(A|E_1) = 9/10$ and $P(A|E_2) = 1/10$. So $P(E_1|A) = (1/100 \cdot 9/10)/(1/100 \cdot 9/10 + 99/100 \cdot 1/10) = 9/108 = 1/12$.

Change the prior

Your turn: say 20 people in 1000 are ill, and the test is still right 90% of the time. How many test positive? What is the chance that a person who tests positive is ill?

First try

(1/3)/(1/3)

1

Both terms

(1/3)/(1/3 + 1/6)

(1/3)/(1/2)

Answer

2/3

A common slip is to leave a case out of the denominator of Bayes’ theorem. Let us see what that does.

Go back to the two bags. Let $A$ be “the coin picks bag A” and $B$ be “the coin picks bag B”. Each has chance $1/2$. Let $R$ be “the ball is red”. Then $P(R|A) = 2/3$ and $P(R|B) = 1/3$. You draw a red ball. How likely is it that it came from bag A?

The top of Bayes’ theorem is $1/2 \cdot 2/3 = 1/3$. The bottom must be all of $P(R)$, from both bags: $P(A|R) = (1/2 \cdot 2/3)/(1/2 \cdot 2/3 + 1/2 \cdot 1/3) = (1/3)/(1/2) = 2/3$.

Now drop bag B’s term from the bottom. You get $(1/3)/(1/3) = 1$. That says the red ball surely came from bag A. But bag B holds a red ball too, so that answer is wrong.

The figure draws the bags as urns. Its strip shows four cases. They are urn A red, urn A white, urn B red and urn B white. Their widths are $1/3$, $1/6$, $1/6$ and $1/3$. The two red cases add to $1/2$. With both terms, $P(A|R) = 2/3$. With urn B dropped, it jumps to 1.

The denominator of Bayes’ theorem sums over every case of the partition. That includes the cases you are not asking about.

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Turning outcomes into numbers

A stall at the school fair has a game. You throw two dice. If the total is 12, you win Rs 50. If the total is 7, you win Rs 10. Any other total wins nothing. How much does one throw win?

It depends on the throw. But each throw gives one amount. The 36 outcomes of the two dice turn into just three amounts: Rs 50, Rs 10 and Rs 0.

A RANDOM VARIABLE is a function that gives a real number to each outcome of a random experiment. Here $X$ is the amount you win. For the outcome $(6, 6)$, $X = 50$. For $(3, 4)$, $X = 10$. For $(2, 2)$, $X = 0$.

The PROBABILITY DISTRIBUTION of $X$ lists each value $x_i$ with its chance $p_i$. Only $(6, 6)$ gives a total of 12, so $P(X = 50) = 1/36$. Six pairs make 7, so $P(X = 10) = 6/36$. The other 29 outcomes give $P(X = 0) = 29/36$.

A list like this is a probability distribution only if two things hold. - Each $p_i$ is between 0 and 1. - The $p_i$ add up to 1. Here $1/36 + 6/36 + 29/36 = 36/36 = 1$.

The figure checks both on four values. $X$ takes the values 1 to 4. Their chances are $1/10$, $3/10$, $4/10$ and $2/10$. On the left, each bar stays under the line at 1. On the right, the four are stacked, and they reach 1 exactly.

A list of chances

Your turn: a list gives the chances 0.5, 0.3 and 0.3. Is it a probability distribution?

paid out

kept

Many people play the stall’s game. What does one throw pay, on average? Out of 36 throws, about 1 pays Rs 50 and about 6 pay Rs 10. That is Rs 110 over 36 throws, or about Rs 3.06 a throw.

That average is the MEAN of $X$, also called its EXPECTATION. It is $E(X) = \sum_{i=1}^n x_i p_i$. It is also written $\mu$, so $E(X)$ and $\mu$ are two names for one number. For the stall, $E(X) = 50 \cdot 1/36 + 10 \cdot 6/36 + 0 \cdot 29/36 = 110/36$.

No single throw pays Rs 3.06. The mean is an average over many throws. If the stall charges Rs 5 a throw, it keeps about Rs 1.94 a throw over a long day.

Balancing five values

The five-bar figure shows the mean as a balance point. $X$ runs from 1 to 5. Its chances are 8, 6, 3, 2 and 1 twentieths. The bars balance at $E(X) = 2.1$, between 2 and 3.

SD about 0.89

3

SD about 1.61

3

The mean does not tell you how far the values spread. The VARIANCE does. It is $\operatorname{Var}(X) = \sum_{i=1}^n (x_i - \mu)^2 p_i$.

1. $(x_i - \mu)^2 = x_i^2 - 2 \mu x_i + \mu^2$, by opening the square.
2. $\operatorname{Var}(X) = \sum x_i^2 p_i - 2 \mu \sum x_i p_i + \mu^2 \sum p_i$, adding that term by term.
3. $\sum x_i p_i = \mu$ and $\sum p_i = 1$, because those are exactly $E(X)$ and the total probability.
4. $\operatorname{Var}(X) = E(X^2) - 2 \mu^2 + \mu^2 = E(X^2) - \mu^2$, substituting both facts in.

So $\operatorname{Var}(X) = E(X^2) - \mu^2$, where $E(X^2) = \sum_{i=1}^n x_i^2 p_i$. The STANDARD DEVIATION is the positive square root, $\sqrt{\operatorname{Var}(X)}$.

Same mean, different spread

The two bar charts you compared earlier, both with mean 3, show why the spread matters. On the left, half the values sit at 3, and the variance is 0.8. On the right, most sit at the ends, and the variance is 2.6.

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Updating belief with evidence

The first page asked a question. How does a chance change when you learn something? And how do you work back from what you saw to what most likely caused it? Here is the answer, step by step.

• CONDITIONAL PROBABILITY takes the share of what you know: $P(A|B) = P(A \cap B)/P(B)$, with $P(B) \neq 0$. Given an even number on a die, a 6 has chance $(1/6)/(3/6) = 1/3$.
• The MULTIPLICATION THEOREM builds a chance step by step: $P(A \cap B) = P(A) \cdot P(B|A)$, with $P(A) \neq 0$. Two red sweets from 4 red and 2 blue, with none put back: $2/3 \cdot 3/5 = 2/5$.
• INDEPENDENT EVENTS are the case where learning one changes nothing: $P(A \cap B) = P(A) \cdot P(B)$. A head and a 6: $1/2 \cdot 1/6 = 1/12$. Three events need all four product rules.
• TOTAL PROBABILITY adds over a partition: $P(A) = \sum_i P(E_i) \cdot P(A|E_i)$. The cases are disjoint, they make up $S$, and each has $P(E_i) > 0$. Two bags picked by a coin: $1/2 \cdot 2/3 + 1/2 \cdot 1/3 = 1/2$.
• BAYES’ THEOREM turns this round: $P(E_i|A) = (P(E_i) \cdot P(A|E_i))/P(A)$, with $P(A) \neq 0$. A positive test with 10 ill in 1000: $9/108 = 1/12$.
• A RANDOM VARIABLE puts a number on each outcome. Its chances lie between 0 and 1 and add to 1. Its MEAN is $E(X) = \sum x_i p_i$, and its VARIANCE is $E(X^2) - [E(X)]^2$. Heads in two tosses: mean 1, variance $1/2$. (JEE tier)

The face-down card

Back to the card on the table. A king was 1 chance in 13. Then you heard “It is a picture card”, and the chance became 1 in 3. The card was never turned over. You still do not know what it is; you only know more. Your new chance is the chance of a king, given a picture card. That is a conditional probability. Working back from a positive test to the illness is the same idea, turned round by Bayes’ theorem.

Looking back on the book

This is also the last chapter of the book. Across thirteen chapters you learned to measure how fast a thing changes. You learned to add up a whole from its small parts. You solved equations with matrices and measured lines in space. You found the best plan within fixed limits. And now you can reason when the answer is not certain. Each of these will come back in the next course you take.

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Practice set: Exercise 13.1

Exercise 13.1 practises conditional probability. You count inside the event you are told about, or you divide by its chance. Some items use the complement rule, some draw without putting back, and one asks about $P(B) = 0$.

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Practice set: Exercise 13.2

Exercise 13.2 practises the multiplication theorem and independent events. Some draws put the first ball back, and some do not. You test pairs with the product rule, and you tell independent events from mutually exclusive ones. Some items have three independent events, or ask for “at least one”.

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Practice set: Exercise 13.3

Exercise 13.3 practises total probability and Bayes’ theorem. The cases come from bags, factories, a clinic test, a cricket toss and a bank. First you add over every case to find a chance. Then Bayes’ theorem gives the chance of one case, once you know the outcome.

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Practice set: random variables

This set on random variables practises probability distributions. You check that a list of chances adds to 1, and you find an unknown $k$. You build the distribution of the red balls in a draw, and of the difference on two spinners. Then you find $E(X)$ and, for some, $\operatorname{Var}(X)$.

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Practice set: miscellaneous

The miscellaneous set mixes the ideas, and no item says which one it needs. You test events for independence, two and three at a time, and you combine two independent tests. You find a missing chance from independence. You prove that complements of independent events are independent, and you tell mutually exclusive events from independent ones. You find conditional chances with cards, with complements and for a device of two parts. Bayes’ theorem comes up for a factory, a cricket match and a reported coin toss. One item, for the JEE tier, finds a variance.