Chapter 2: Conditional Probability

Conditional Probability

Definition, Formula and Solved Examples

Definition of Conditional Probability

Let A and B be two events in a sample space S, where

P(B) > 0.

The conditional probability of A given that B has already occurred is denoted by

P(A | B)

and is defined as

P(A | B) = P(A ∩ B) / P(B)

The occurrence of event B reduces the original sample space to B. We then determine what proportion of B also belongs to A.

Figure 1: Conditional probability considers the part of A lying inside B.

The multiplication rule follows directly:

P(A ∩ B) = P(A | B) P(B)

Similarly,

P(A ∩ B) = P(B | A) P(A)

Example 1: Drawing a Card

A card is drawn randomly from a standard deck of 52 cards. Given that the selected card is a face card, find the probability that it is a king.

Figure 2: Kings, Queens and Jacks form the 12 face cards.

Show Solution

Let

A = {the card is a king}

and

B = {the card is a face card}.

There are 12 face cards:

4 Kings + 4 Queens + 4 Jacks = 12.

Therefore,

P(B) = 12/52.

All four kings are face cards, so

P(A ∩ B) = 4/52.

Using conditional probability,

P(A | B) = P(A ∩ B)/P(B)

= (4/52)/(12/52)

= 4/12 = 1/3.

Answer: P(A | B) = 1/3

Example 2: Rolling Two Dice

Two fair dice are rolled. Given that the sum of the two numbers is 8, find the probability that at least one die shows 5.

Figure 3: Five possible ordered outcomes have sum 8.

Show Solution

Let

B = {sum of the two dice is 8}.

The possible outcomes are

(2,6), (3,5), (4,4), (5,3), (6,2).

Hence,

n(B)=5.

Let

A = {at least one die shows 5}.

The outcomes satisfying both conditions are

(3,5), (5,3).

Therefore,

n(A ∩ B)=2.

Thus,

P(A | B)=2/5.

Answer: 2/5

Example 3: Mathematics and Physics Students

In a class of 100 students, 60 study Mathematics, 45 study Physics and 30 study both subjects. If a randomly selected student studies Mathematics, find the probability that the student also studies Physics.

Figure 4: Venn representation of Mathematics and Physics students.

Show Solution

Let

M = {student studies Mathematics}

and

P = {student studies Physics}.

We have

P(M)=60/100

and

P(M ∩ P)=30/100.

Therefore,

P(P | M) = P(P ∩ M)/P(M)

= (30/100)/(60/100)

= 30/60 = 1/2.

Answer: 1/2 = 0.5 = 50%

Example 4: Defective Components

A factory produces electronic components. It is known that 8% of the components are defective and 3% are both defective and produced during the night shift. If a component is known to be defective, find the probability that it was produced during the night shift.

Figure 5: The 3% night-shift defective group lies inside the 8% defective group.

Show Solution

Let

D = {component is defective}

and

N = {component was produced during night shift}.

Given

P(D)=0.08

and

P(D ∩ N)=0.03.

We require

P(N | D).

Using the definition,

P(N | D) = P(N ∩ D)/P(D)

= 0.03/0.08

= 0.375.

Answer: 0.375 = 37.5%

Example 5: Selection Without Replacement

A box contains 5 red balls and 3 blue balls. Two balls are drawn successively without replacement. Find the probability that the second ball is red, given that the first ball is red.

Figure 6: After the first red ball is removed, 7 balls remain.

Show Solution

Originally there are

5+3=8

balls.

Since the first ball is already known to be red, one red ball has been removed.

The remaining balls are

4 red + 3 blue = 7 balls.

Therefore,

P(second red | first red) = 4/7.

Answer: 4/7

Example 6: Random Integer

An integer is selected randomly from {1,2,3,...,30}. Given that the integer is divisible by 3, find the probability that it is also divisible by 6.

Multiples of 3:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Figure 7: The bold numbers are divisible by both 3 and 6.

Show Solution

Let

A = {integer is divisible by 6}

and

B = {integer is divisible by 3}.

The multiples of 3 are

3,6,9,12,15,18,21,24,27,30.

Hence

n(B)=10.

The numbers divisible by 6 are

6,12,18,24,30.

Therefore

n(A ∩ B)=5.

Thus,

P(A | B) = 5/10 = 1/2.

Answer: 1/2

Example 7: Conditional Probability from a Table

A survey of 200 university students gives the following information. If a randomly selected student has passed, find the probability that the student is female.

	Passed	Failed	Total
Male	80	20	100
Female	70	30	100
Total	150	50	200

Figure 8: Contingency table for gender and examination result.

Show Solution

Let

F = {student is female}

and

P = {student passed}.

From the table,

n(P)=150

and

n(F ∩ P)=70.

Therefore,

P(F | P) = 70/150

= 7/15.

In decimal form,

7/15 ≈ 0.4667.

Answer: 7/15 ≈ 46.67%

Example 8: Three Coin Tosses

A fair coin is tossed three times. Given that at least one head occurs, find the probability that exactly two heads occur.

Figure 9: Tree representation for three tosses of a coin.

Show Solution

The sample space is

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}.

Let

B = {at least one head occurs}.

The only outcome with no head is TTT. Therefore,

B= {HHH,HHT,HTH,HTT,THH,THT,TTH}.

Hence,

n(B)=7.

Exactly two heads occur in

HHT, HTH, THH.

Thus,

n(A ∩ B)=3.

Therefore,

P(A | B)=3/7.

Answer: 3/7

Example 9: Medical Diagnostic Test and Bayes' Theorem

Suppose

P(D)=0.02,

where D denotes that a person has a disease.

The diagnostic test satisfies

P(+ | D)=0.95

and

P(+ | Dᶜ)=0.05.

Find the probability that a person actually has the disease given that the test result is positive.

Figure 10: Probability tree for the medical test.

Show Solution

We require

P(D | +).

By Bayes' theorem,

P(D | +) = [P(+ | D)P(D)] / P(+).

First calculate the probability of obtaining a positive test:

P(+) = P(+ | D)P(D) + P(+ | Dᶜ)P(Dᶜ).

Since

P(Dᶜ)=1-0.02=0.98,

we have

P(+) = (0.95)(0.02) + (0.05)(0.98).

= 0.019+0.049 = 0.068.

Therefore,

P(D | +) = (0.95)(0.02)/0.068.

= 0.019/0.068.

≈0.2794.

Answer: P(D | +) ≈ 0.2794 = 27.94%

A positive test does not automatically imply a very high probability of disease. The low prevalence of the disease affects the posterior probability.

Example 10: Three Events and the Multiplication Rule

Suppose events A, B and C satisfy

P(A)=0.6,

P(B | A)=0.5,

P(C | A ∩ B)=0.4.

Find

P(A ∩ B ∩ C).

Figure 11: Successive conditional probabilities in the multiplication rule.

Show Solution

For three events, the multiplication rule is

P(A ∩ B ∩ C) = P(A) P(B | A) P(C | A ∩ B).

Substituting the given values,

P(A ∩ B ∩ C) = (0.6)(0.5)(0.4).

= 0.12.

Answer: P(A ∩ B ∩ C)=0.12

Important Formulas

1. Conditional Probability

P(A | B) = P(A ∩ B) / P(B),    P(B) > 0

2. Reverse Conditional Probability

P(B | A) = P(A ∩ B) / P(A),    P(A) > 0

3. Multiplication Rule

P(A ∩ B) = P(A)P(B | A)

4. Equivalent Multiplication Rule

P(A ∩ B) = P(B)P(A | B)

5. Three-Event Multiplication Rule

P(A ∩ B ∩ C) = P(A) P(B | A) P(C | A ∩ B)

6. Independent Events

If A and B are independent,

P(A | B)=P(A)

and therefore

P(A ∩ B) = P(A)P(B)

7. Bayes' Theorem

P(A | B) = [P(B | A)P(A)] / P(B)

Conditional Probability — University Level Mathematics
Practice the examples first and then click Show Solution to check your answer.