Conditional probability

A school surveys $200$ grade-12 students about joining a sports club:

	Joined	Not joined	Total
Boys	$60$	$40$	$100$
Girls	$30$	$70$	$100$
Total	$90$	$110$	$200$

One student is chosen at random. The probability that the student has joined the club is $\dfrac{90}{200}=0.45$ . But if we also know that the student chosen is a girl, what is the probability that she has joined the club?

Explore: shrinking the sample space

Once we know that the student is a girl, we only look at the $100$ girls in the second row. Of these, $30$ have joined, so the probability we want is

$\frac{30}{100}=0.3.$

This is not $0.45$ : the information "is a girl" has changed the probability. We can also write it with probabilities over all $200$ students:

$\frac{30}{100}=\frac{30/200}{100/200}=\frac{P(\text{girl and joined})}{P(\text{girl})}=\frac{0.15}{0.5}=0.3.$

Why this formula? Knowing that $B$ has happened means only the outcomes in $B$ are still possible, so $B$ acts as a new sample space. Inside it, the outcomes that make $A$ happen are exactly $A\cap B$ . We divide by $P(B)$ so that the probabilities in the "new space" add up to $1$ .

The definition gives $P(\overline{A}\mid B)=1-P(A\mid B)$ , just as for ordinary probabilities.

The order of the two events in $P(A\mid B)$ matters a great deal, as the next example shows.

The multiplication rule and tree diagrams

Multiplying both sides of the definition by $P(B)$ gives a formula for the probability of an intersection.

This rule is especially useful when an experiment happens in stages. On a tree diagram, each branch carries the conditional probability of that stage, and the probability of a path is the product of the numbers along it.

Tree diagram for drawing two balls without replacement

The four numbers at the ends add up to $\dfrac{20+15+15+6}{56}=1$ , a quick check.

Independent events

In grade 11, two events were called independent when whether one happens does not change the probability of the other. With conditional probability we can write this as a formula: when $P(B)>0$ ,

$A \text{ and } B \text{ are independent}\iff P(A\mid B)=P(A)\iff P(A\cap B)=P(A)\cdot P(B).$

The second equivalence follows from the multiplication rule. In the drawing example, $P(D_2\mid D_1)=\dfrac47$ but $P(D_2\mid\overline{D_1})=\dfrac57$ , so $D_1$ and $D_2$ are not independent.

The last example is a medical one, where mixing up $P(A\mid B)$ and $P(B\mid A)$ can cause serious misunderstanding.

Example Screening test results

A screening programme tests $1000$ people, with these results:

	Has the disease	Does not	Total
Positive	$45$	$95$	$140$
Negative	$5$	$855$	$860$
Total	$50$	$950$	$1000$

One person is chosen at random. Find the probability of a positive test given that the person has the disease, and the probability of having the disease given a positive test.

Solution. The first question only uses the "Has the disease" column, the second only the "Positive" row:

$P(\text{positive}\mid\text{disease})=\frac{45}{50}=0.9,\qquad P(\text{disease}\mid\text{positive})=\frac{45}{140}=\frac{9}{28}\approx0.32.$

The test detects $90\%$ of the people who have the disease, but only about $32\%$ of those who test positive really have it, because the healthy people are so many that the $95$ false positives outnumber the $45$ true ones. The next lesson explains this with Bayes' theorem.

Watch out Common mistakes

Mixing up $P(A\mid B)$ and $P(B\mid A)$ . Wrong: "the test is right for $90\%$ of sick people, so a person who tests positive is $90\%$ likely to be sick". Why it is wrong: that is $P(\text{positive}\mid\text{disease})$ , not $P(\text{disease}\mid\text{positive})$ . Right: $P(\text{disease}\mid\text{positive})=\dfrac{45}{140}\approx0.32$ .
Dividing by the whole sample space. Wrong: in the dice example, $P(A\mid B)=\dfrac{2}{36}$ . Why it is wrong: once $B$ is known, only $5$ outcomes remain. Right: $\dfrac25$ .
The wrong denominator. Wrong: $P(A\mid B)=\dfrac{P(A\cap B)}{P(A)}$ . Why it is wrong: the condition is $B$ , so we divide by $P(B)$ .
Multiplying as if independent when drawing without replacement. Wrong: $P(\text{two red})=\dfrac58\cdot\dfrac58=\dfrac{25}{64}$ . Why it is wrong: after the first draw the box has changed. Right: $\dfrac58\cdot\dfrac47=\dfrac5{14}$ .

Key terms

English	Tiếng Việt
conditional probability	xác suất có điều kiện
intersection of events	biến cố giao
two-way table	bảng hai chiều
tree diagram	sơ đồ cây
multiplication rule	công thức nhân xác suất
independent events	biến cố độc lập
drawing without replacement	lấy không hoàn lại
false positive	dương tính giả

Summary

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ (with $P(B)>0$ ); for equally likely outcomes, $\dfrac{n(A\cap B)}{n(B)}$ .
Knowing that $B$ happened shrinks the sample space to $B$ ; in a two-way table, read only the row or column of $B$ .
$P(A\mid B)$ and $P(B\mid A)$ are usually different.
Multiplication rule: $P(A\cap B)=P(B)\cdot P(A\mid B)$ ; on a tree diagram, multiply along a path.
$A$ , $B$ independent $\iff P(A\mid B)=P(A)\iff P(A\cap B)=P(A)P(B)$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

A fair die is rolled. Given that the result is odd, find the probability that it is at most $3$ .

What you need and how to start

What you need: $P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}$ when outcomes are equally likely.

Steps:

List $B=\{1, 3, 5\}$ .
Count the elements of $B$ that are at most $3$ .

Show solution

$A\cap B=\{1, 3\}$ , so $P(A\mid B)=\dfrac23$ .

Answer. $\dfrac23$ .

Exercise 2

A class has $40$ students: $20$ boys, $14$ of whom like maths, and $20$ girls, $10$ of whom like maths. One student is chosen at random. Find the probability that the student likes maths given that she is a girl, and the probability that the student is a boy given that the student likes maths.

What you need and how to start

What you need: make a two-way table; read only the row (or column) of the condition.

Steps:

The number who like maths is $14+10=24$ .
Compute $\dfrac{10}{20}$ and $\dfrac{14}{24}$ .

Show solution

$P(\text{likes maths}\mid\text{girl})=\dfrac{10}{20}=\dfrac12$ ; $P(\text{boy}\mid\text{likes maths})=\dfrac{14}{24}=\dfrac{7}{12}$ .

Answer. $\dfrac12$ and $\dfrac7{12}$ .

Exercise 3

Given $P(A)=0.5$ , $P(B)=0.4$ and $P(A\cap B)=0.2$ , find $P(A\mid B)$ and $P(B\mid A)$ , and decide whether $A$ and $B$ are independent.

What you need and how to start

What you need: the definition of conditional probability; independence means $P(A\mid B)=P(A)$ .

Steps:

Divide $P(A\cap B)$ by $P(B)$ , then by $P(A)$ .
Compare $P(A\mid B)$ with $P(A)$ .

Show solution

$P(A\mid B)=\dfrac{0.2}{0.4}=0.5$ and $P(B\mid A)=\dfrac{0.2}{0.5}=0.4$ . Since $P(A\mid B)=P(A)$ , $A$ and $B$ are independent.

Answer. $0.5$ ; $0.4$ ; independent.

Exercise 4

A box holds $4$ white balls and $6$ black balls. Two balls are drawn one after the other without replacement. Find the probability that both are black, and the probability that the second is white given that the first is black.

What you need and how to start

What you need: the multiplication rule $P(A\cap B)=P(A)\cdot P(B\mid A)$ ; after the first draw, $9$ balls remain.

Steps:

$P(\text{black first})=\dfrac{6}{10}$ and $P(\text{black second}\mid\text{black first})=\dfrac59$ .
If the first is black, $4$ white remain out of $9$ .

Show solution

$P(\text{both black})=\dfrac{6}{10}\cdot\dfrac59=\dfrac13$ . $P(\text{white second}\mid\text{black first})=\dfrac49$ .

Answer. $\dfrac13$ and $\dfrac49$ .

Exercise 5

A box holds $10$ cards numbered $1$ to $10$ . One card is drawn at random. Let $A$ be "the number is even" and $B$ be "the number is at least $7$ ". Is each claim true or false?

a) $P(B)=0.4$ .

b) $P(A\mid B)=\dfrac12$ .

c) $P(B\mid A)=\dfrac25$ .

d) $A$ and $B$ are not independent.

What you need and how to start

What you need: $P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}$ ; independence means $P(A\mid B)=P(A)$ .

Steps:

List $B=\{7, 8, 9, 10\}$ and $A\cap B=\{8, 10\}$ .
Compare $P(A\mid B)$ with $P(A)=\dfrac12$ .

Show solution

a) True: $P(B)=\dfrac4{10}=0.4$ .

b) True: $P(A\mid B)=\dfrac24=\dfrac12$ .

c) True: $A$ has $5$ elements, of which $8$ and $10$ are in $B$ , so $P(B\mid A)=\dfrac25$ .

d) False: $P(A\mid B)=\dfrac12=P(A)$ , so $A$ and $B$ are independent.

Answer. a) True; b) True; c) True; d) False.

Exercise 6

Two fair dice are rolled. Given that at least one die shows a $6$ , find the probability that the total is at least $10$ . Give your answer to 2 decimal places.

What you need and how to start

What you need: counting in the shrunken sample space.

Steps:

Count the outcomes with at least one $6$ : $36-25=11$ .
Among them, count those with total $\ge10$ .

Show solution

The outcomes with a $6$ and total $\ge10$ are $(4, 6)$ , $(6, 4)$ , $(5, 6)$ , $(6, 5)$ , $(6, 6)$ . The probability is $\dfrac5{11}\approx0.4545$ .

Answer. $0.45$ .

Exercise 7

In a café, $70\%$ of customers order a coffee drink, and $40\%$ of those also order a cake. Also, $35\%$ of all customers order a cake. Find the probability that a customer orders both coffee and cake, and the probability that a customer ordered coffee given that the customer ordered cake.

What you need and how to start

What you need: " $40\%$ of those" is the conditional probability $P(\text{cake}\mid\text{coffee})=0.4$ .

Steps:

Use the multiplication rule.
Divide by $P(\text{cake})=0.35$ .

Show solution

$P(\text{coffee and cake})=0.7\cdot0.4=0.28$ . $P(\text{coffee}\mid\text{cake})=\dfrac{0.28}{0.35}=0.8$ .

Answer. $0.28$ and $0.8$ .

Exercise 8 Review

Two marksmen fire independently at a target, one shot each, hitting with probabilities $0.8$ and $0.7$ . Given that the target is hit at least once, find the probability that both hit it. Give your answer to 2 decimal places.

What you need and how to start

What you need: the multiplication rule for independent events (grade 11) and the complementary event: $P(\text{at least one})=1-P(\text{nobody hits})$ .

Steps:

$P(\text{both hit})=0.8\cdot0.7$ .
$P(\text{at least one hit})=1-0.2\cdot0.3$ ; "both hit" lies inside "at least one hit".

Show solution

$P(\text{both})=0.56$ and $P(\text{at least one})=1-0.06=0.94$ . Since "both hit" implies "at least one hit", their intersection is "both hit", so the probability is $\dfrac{0.56}{0.94}=\dfrac{28}{47}\approx0.596$ .

Answer. $\approx0.60$ .

Exercise 9 Review

A group has $5$ boys and $4$ girls. Three of them are chosen at random. Given that at least one girl is chosen, find the probability that exactly one girl is chosen.

What you need and how to start

What you need: counting with combinations (see permutations, arrangements and combinations).

Steps:

The number of choices with at least one girl is $\binom{9}{3}-\binom{5}{3}$ .
The number with exactly one girl is $\binom{4}{1}\cdot\binom{5}{2}$ .

Show solution

$\binom{9}{3}-\binom{5}{3}=84-10=74$ and $\binom{4}{1}\cdot\binom{5}{2}=4\cdot10=40$ . The probability is $\dfrac{40}{74}=\dfrac{20}{37}$ .

Answer. $\dfrac{20}{37}$ .

Exercise 10 Challenge

A fair coin is tossed three times. Let $A$ be "the first toss is heads" and $B$ be "at least two tosses are heads". Find $P(A\mid B)$ and $P(B\mid A)$ , and decide whether $A$ and $B$ are independent.

What you need and how to start

What you need: list the $8$ equally likely outcomes; independence means $P(A\mid B)=P(A)$ .

Steps:

Write $B$ (H for heads, T for tails): HHT, HTH, THH, HHH.
Count $A\cap B$ , then divide by $n(B)$ and by $n(A)=4$ .

Show solution

$A\cap B=\{\text{HHT}, \text{HTH}, \text{HHH}\}$ has $3$ elements. $P(A\mid B)=\dfrac34$ and $P(B\mid A)=\dfrac34$ . Since $P(A)=\dfrac12\neq\dfrac34$ , the events are not independent: knowing the first toss is heads makes at least two heads more likely.

Answer. $\dfrac34$ ; $\dfrac34$ ; not independent.

Conditional probability