Conditional probability

What you will learn

calculate conditional probabilities using $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ ,
interpret the language of conditional probability: “given”, “if”, “of”, “knowing that”,
determine whether two events are independent,
apply conditional probability to sampling without replacement,
read and construct two-way tables to find conditional probabilities,
draw tree diagrams with conditional branches for dependent events.

Worked example 0 Real-world example: medical testing

A screening test for a rare disease has the following properties: the disease affects $1\%$ of the population. If a person has the disease, the test is positive $95\%$ of the time. If a person does not have the disease, the test is positive $3\%$ of the time (false positive). What is the probability that a person who tests positive actually has the disease?

Let $D$ = has disease, $T^+$ = tests positive.
$P(D) = 0.01$ , $P(T^+ \mid D) = 0.95$ , $P(T^+ \mid \overline{D}) = 0.03$ .
$P(T^+) = P(T^+ \mid D) \cdot P(D) + P(T^+ \mid \overline{D}) \cdot P(\overline{D}) = 0.95 \times 0.01 + 0.03 \times 0.99 = 0.0095 + 0.0297 = 0.0392$ .
$P(D \mid T^+) = \dfrac{P(T^+ \mid D) \cdot P(D)}{P(T^+)} = \dfrac{0.0095}{0.0392} \approx 0.242$ .

Key idea: even with an accurate test, only about $24\%$ of positive results are true positives when the disease is rare. This is why confirmatory testing exists.

1. Conditional probability

The conditional probability of event $A$ given event $B$ has occurred is:

Definition

$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}, \quad P(B) > 0$

Read $P(A \mid B)$ as “the probability of $A$ given $B$ .”

The vertical bar " $\mid$ " means “given that” or “knowing that.” Watch for these phrases in word problems:

“If a student plays sport, what is the probability they also play music?” means $P(\text{music} \mid \text{sport})$ .
“Of those who passed, what fraction studied more than $5$ hours?” means $P(\text{studied} > 5 \mid \text{passed})$ .

Worked example 1 Using the formula

In a group of $40$ students, $25$ study maths, $18$ study science, and $10$ study both. Find $P(\text{science} \mid \text{maths})$ .

$P(\text{maths}) = \dfrac{25}{40}$ .
$P(\text{science and maths}) = \dfrac{10}{40}$ .
$P(\text{science} \mid \text{maths}) = \dfrac{P(\text{science and maths})}{P(\text{maths})} = \dfrac{10/40}{25/40} = \dfrac{10}{25} = \dfrac{2}{5} = 0.4$ .

Interpretation: of the $25$ maths students, $10$ also study science — that is $40\%$ .

2. Independent and dependent events

Two events are independent if knowing one occurred does not change the probability of the other:

$P(A \mid B) = P(A) \quad \text{(independence test)}$

Equivalently, for independent events: $P(A \text{ and } B) = P(A) \times P(B)$ .

If $P(A \mid B) \neq P(A)$ , the events are dependent.

Worked example 2 Testing for independence

From the example above: $P(\text{science}) = \dfrac{18}{40} = 0.45$ and $P(\text{science} \mid \text{maths}) = 0.4$ .

Since $0.4 \neq 0.45$ , the events “studies science” and “studies maths” are dependent. Knowing a student studies maths slightly decreases the probability they study science.

3. Two-way tables for conditional probability

A two-way table provides all the information needed to calculate conditional probabilities directly from counts.

Worked example 3 Conditional probability from a two-way table

A survey of $200$ adults records exercise habits and health ratings:

	Good health	Poor health	Total
Exercises regularly	90	30	120
Does not exercise	40	40	80
Total	130	70	200

(a) $P(\text{good health} \mid \text{exercises}) = \dfrac{90}{120} = \dfrac{3}{4} = 0.75$ .

(b) $P(\text{good health} \mid \text{does not exercise}) = \dfrac{40}{80} = \dfrac{1}{2} = 0.5$ .

Note: $P(A \mid B) \neq P(B \mid A)$ in general. Part (a) and part (c) have different values.

4. Tree diagrams with conditional branches

When events are dependent, the branches of a tree diagram show conditional probabilities. This is especially useful for sampling without replacement.

Tree diagram: drawing 2 cards from a hand of 3 red and 2 black cards without replacement.

The second-draw probabilities are conditional on the first draw’s result. After drawing a red card first, only $2$ red and $2$ black remain out of $4$ total.

Worked example 4 Using the tree diagram

From the tree diagram above, find:

(a) $P(\text{at least one red}) = P(RR) + P(RB) + P(BR) = \dfrac{6}{20} + \dfrac{6}{20} + \dfrac{6}{20} = \dfrac{18}{20} = \dfrac{9}{10}$ .

(b) $P(\text{2nd card is red} \mid \text{1st card is black}) = \dfrac{3}{4}$ .

This can be read directly from the tree: on the lower branch (1st card black), the “R” branch has probability $\dfrac{3}{4}$ .

Practice

Fluency

Tier 1: basic skills

In a class of $30$ students, $12$ play basketball. What is $P(\text{basketball})$ ?
Of the $12$ basketball players, $5$ are also in the swim team. What is $P(\text{swim} \mid \text{basketball})$ ?
A bag has $6$ red and $4$ blue marbles. One marble is drawn and not replaced. If the first marble was red, what is $P(\text{red on 2nd draw})$ ?
Events $A$ and $B$ satisfy $P(A) = 0.3$ , $P(B) = 0.5$ , $P(A \text{ and } B) = 0.15$ . Find $P(A \mid B)$ .
Are $A$ and $B$ in Q4 independent? Justify.
A two-way table shows: $20$ males own a pet, $15$ males do not, $25$ females own a pet, $10$ females do not. Find $P(\text{pet} \mid \text{male})$ .
Using the same table, find $P(\text{male} \mid \text{pet})$ .
A coin is tossed and a die is rolled. Are the events “heads” and “rolling a 6” independent? Explain.
Two cards are drawn without replacement from a deck of $52$ . Find $P(\text{2nd card is heart} \mid \text{1st card is heart})$ .
State the formula for $P(A \mid B)$ .

Reasoning

Tier 2: mixed practice

A box contains $5$ green and $3$ yellow balls. Two balls are drawn without replacement. Draw a tree diagram with conditional probabilities on each branch, and find $P(\text{both green})$ .
In a school of $400$ students, $240$ study French, $180$ study German, and $80$ study both. Find: (a) $P(\text{German} \mid \text{French})$ , (b) $P(\text{French} \mid \text{German})$ .
A survey finds:

Supports policy Opposes policy Total
Under 30 45 30 75
30 and over 35 40 75
Total 80 70 150

(a) Find $P(\text{supports} \mid \text{under 30})$ and $P(\text{supports} \mid \text{30 and over})$ . (b) Is there an association between age group and opinion? Justify.
Events $A$ and $B$ are such that $P(A) = 0.6$ , $P(B) = 0.4$ , and $P(A \mid B) = 0.75$ . Find $P(A \text{ and } B)$ and determine whether $A$ and $B$ are independent.
Three machines produce items. Machine X makes $50\%$ of items with a $2\%$ defect rate. Machine Y makes $30\%$ with a $3\%$ defect rate. Machine Z makes $20\%$ with a $5\%$ defect rate. An item is selected at random. Find the probability it is defective.

	Supports policy	Opposes policy	Total
Under 30	45	30	75
30 and over	35	40	75
Total	80	70	150

Reasoning

Tier 3: explain and apply

Using the machine data from Tier 2 Q5, an item is found to be defective. Find the probability it came from Machine Z.
Explain, with a numerical example, why $P(A \mid B) \neq P(B \mid A)$ in general. Why is confusing these two a common and dangerous error in medical or legal contexts?
A jar contains $4$ red and $6$ blue marbles. Three marbles are drawn without replacement. Find $P(\text{all three are blue})$ using a chain of conditional probabilities.
Two events satisfy $P(A \mid B) = 0.5$ and $P(B \mid A) = 0.25$ . If $P(B) = 0.4$ , find $P(A)$ .

Challenge

Reasoning

Harder reasoning

A game show has three doors. Behind one door is a prize; behind the other two, nothing. You pick a door. The host, who knows what is behind each door, opens a different door to reveal no prize. You are offered the chance to switch. Using conditional probability, show that switching gives you a $\dfrac{2}{3}$ chance of winning.
In a population, $0.5\%$ use a certain drug. A drug test has a $99\%$ true positive rate and a $2\%$ false positive rate. Find the probability that a person who tests positive actually uses the drug. Comment on the usefulness of the test.
Prove that if $A$ and $B$ are independent, then $A$ and $\overline{B}$ (the complement of $B$ ) are also independent.
Five cards are dealt from a standard deck of $52$ . Find the probability that all five are spades, using a chain of conditional probabilities.

Answers

Answer key

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Tier 1

$P(\text{basketball}) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4$ .
$P(\text{swim} \mid \text{basketball}) = \dfrac{5}{12} \approx 0.417$ .
After removing a red marble: $5$ red and $4$ blue remain out of $9$ . $P(\text{red on 2nd draw}) = \dfrac{5}{9}$ .
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{0.15}{0.5} = 0.3$ .
Yes, $A$ and $B$ are independent because $P(A \mid B) = 0.3 = P(A)$ . Knowing $B$ occurred does not change the probability of $A$ .
$P(\text{pet} \mid \text{male}) = \dfrac{20}{20 + 15} = \dfrac{20}{35} = \dfrac{4}{7} \approx 0.571$ .
$P(\text{male} \mid \text{pet}) = \dfrac{20}{20 + 25} = \dfrac{20}{45} = \dfrac{4}{9} \approx 0.444$ .
Yes, they are independent. The outcome of the coin does not affect the die, and vice versa. $P(\text{heads and 6}) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12} = P(\text{heads}) \times P(\text{6})$ .
After removing one heart, $12$ hearts remain out of $51$ cards. $P(\text{2nd heart} \mid \text{1st heart}) = \dfrac{12}{51} = \dfrac{4}{17}$ .
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ , where $P(B) > 0$ .

Tier 2

First draw: $P(G) = \dfrac{5}{8}$ , $P(Y) = \dfrac{3}{8}$ . If 1st is green: $P(G) = \dfrac{4}{7}$ , $P(Y) = \dfrac{3}{7}$ . If 1st is yellow: $P(G) = \dfrac{5}{7}$ , $P(Y) = \dfrac{2}{7}$ . $P(\text{both green}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
(a) $P(\text{German} \mid \text{French}) = \dfrac{80}{240} = \dfrac{1}{3} \approx 0.333$ . (b) $P(\text{French} \mid \text{German}) = \dfrac{80}{180} = \dfrac{4}{9} \approx 0.444$ .
(a) $P(\text{supports} \mid \text{under 30}) = \dfrac{45}{75} = 0.6$ . $P(\text{supports} \mid \text{30 and over}) = \dfrac{35}{75} \approx 0.467$ . (b) Yes, there is an association. The conditional probabilities differ: younger respondents are more likely to support the policy ( $60\%$ vs $47\%$ ). If there were no association, both groups would have the same support rate of $\dfrac{80}{150} \approx 0.533$ .
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.75 \times 0.4 = 0.3$ . For independence: $P(A) \times P(B) = 0.6 \times 0.4 = 0.24$ . Since $0.3 \neq 0.24$ , $A$ and $B$ are not independent.
$P(\text{defective}) = 0.50 \times 0.02 + 0.30 \times 0.03 + 0.20 \times 0.05 = 0.010 + 0.009 + 0.010 = 0.029$ . The probability that a randomly selected item is defective is $2.9\%$ .

Tier 3

$P(\text{Z and defective}) = 0.20 \times 0.05 = 0.010$ . $P(\text{Z} \mid \text{defective}) = \dfrac{P(\text{Z and defective})}{P(\text{defective})} = \dfrac{0.010}{0.029} \approx 0.345$ . There is about a $34.5\%$ chance the defective item came from Machine Z.
Example: $P(\text{disease} \mid \text{positive test}) \neq P(\text{positive test} \mid \text{disease})$ . A test might detect $95\%$ of sick people ( $P(T^+ \mid D) = 0.95$ ), but if the disease is rare, $P(D \mid T^+)$ could be much lower (e.g. $0.16$ ). Confusing the two — called the “prosecutor’s fallacy” in legal contexts — leads to wildly wrong conclusions. In medicine, it means overestimating how likely a patient is to have the disease after a positive test.
$P(\text{1st blue}) = \dfrac{6}{10}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{5}{9}$ . $P(\text{3rd blue} \mid \text{first two blue}) = \dfrac{4}{8} = \dfrac{1}{2}$ . $P(\text{all three blue}) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{1}{2} = \dfrac{30}{180} = \dfrac{1}{6}$ .
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.4 = 0.2$ . Also $P(A \text{ and } B) = P(B \mid A) \times P(A)$ , so $0.2 = 0.25 \times P(A)$ , giving $P(A) = \dfrac{0.2}{0.25} = 0.8$ .

Challenge

Label the doors $1, 2, 3$ . Suppose the prize is behind door $1$ (by symmetry, the argument is the same for any door). You pick door $1$ : host opens door $2$ or $3$ ; switching loses. You pick door $2$ : host must open door $3$ ; switching wins. You pick door $3$ : host must open door $2$ ; switching wins. So switching wins $2$ out of $3$ times. Formally: let $W$ = prize behind your door. $P(W) = \dfrac{1}{3}$ . Given the host reveals a losing door, $P(\text{prize behind other door} \mid \text{host reveals}) = \dfrac{2}{3}$ .
$P(\text{uses drug}) = 0.005$ . $P(T^+ \mid \text{user}) = 0.99$ . $P(T^+ \mid \text{non-user}) = 0.02$ . $P(T^+) = 0.005 \times 0.99 + 0.995 \times 0.02 = 0.00495 + 0.0199 = 0.02485$ . $P(\text{user} \mid T^+) = \dfrac{0.00495}{0.02485} \approx 0.199$ . Only about $20\%$ of positive results are true positives. The test produces many false positives because the user base is so small. A confirmatory (more specific) test is essential.
$P(A \text{ and } \overline{B}) = P(A) - P(A \text{ and } B) = P(A) - P(A) \cdot P(B)$ (using independence) $= P(A)(1 - P(B)) = P(A) \cdot P(\overline{B})$ . Since $P(A \text{ and } \overline{B}) = P(A) \cdot P(\overline{B})$ , events $A$ and $\overline{B}$ are independent.
$P(\text{all 5 spades}) = \dfrac{13}{52} \times \dfrac{12}{51} \times \dfrac{11}{50} \times \dfrac{10}{49} \times \dfrac{9}{48} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49 \times 48} = \dfrac{154440}{311875200} = \dfrac{33}{66640} \approx 0.000495$ .

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