Formula Name	Formula
Conditional Probability in terms of Probability	P (X/Y) = P (X ∩ Y) / P(Y)
Conditional Probability in terms of Number of Elements	P (X/Y) = n (X ∩ Y) / n(Y)
Conditional Probability Property	P (Xc/Y) = P (X /Y)

Conditional Probability Practice Question with Solution

These Conditional Probability Practice Question with Solution will help you understand the application and use of various formulas on Conditional Probability in different mathematical problems.

1: If P(A) = 0.3, P(B) = 0.7 and P(A∩B) = 0.1 then find P(A/B) and P(B / A).

Given: P(A) = 0.3, P(B) = 0.7 and P(A∩B) = 0.1

Thus, P(A / B) = P(A∩B) / P(B)

⇒ P(A / B) = 0.1 / 0.7

⇒ P(A / B) = 1 / 7

and P(B / A) = P(A∩B) / P(A)

⇒ P(B / A) = 0.1 / 0.3

⇒ P(B / A) = 1 / 3

2: If P(A) = 0.2, P(B/A) = 0.8 and P(A∪B) = 0.3 then find P(B).

Given: P(A) = 0.2, P(B/A) = 0.8 and P(A∪B) = 0.3

We know, P(A∩B) = P(B / A) × P(A)

⇒ P(A∩B) = 0.2 × 0.8

⇒ P(A∩B) = 0.16

and P(A∩B) = P(A) + P(B) – P(A∪B)

⇒ P(B) = P(A∩B) – P(A) + P(A∪B)

⇒ P(B) = 0.16 – 0.2 + 0.3

⇒ P(B) = 0.26

3: A dice is rolled. If X = {1, 2, 6}, Y = {2, 4} and Z = {2, 4, 6} then, find P(X/Y), P(X/Z).

Given: X = {1, 2, 6}, Y = {2, 4} and Z = {2, 4, 6}

Thus, n(X) = 3, n(Y) = 2, n(Z) = 3

and n(X ∩ Y) = 1, n(X ∩ Z) = 2

Using Formula P(X / Y) = n(X ∩ Y) / n(Y)

⇒ P(X / Y) = 1 / 2

and P(X / Z) = n(X ∩ Z) / n(Z)

Thus, P(X / Z) = 2 / 3

4: A coin is tossed 3 times. Find the conditional probability P(C/D) where C = Head on first Toss and D = Tail on second toss

Sample space of three coin toss is

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

n(C) = 4, n(D) = 4

Thus, P(C) = 4/ 8 = 1/2

and P(D) = 4/ 8 = 1/2

Using Formula, P(C ∩ D) = 2 / 8 = 1 /4

⇒ P(C /D) = P(C ∩ D) / P(D)

⇒ P(C /D) = (1 / 4) / (1/2)

Thus, P(C / D) = 1 /2

5: Given that P(A/B) = 3/ 7 then, find the conditional probability P(A^c /B).

By the property of conditional probability

P(A^c / B) = 1 – P(A/B)

⇒ P(A^c / B) = 1 – 3/7

Thus, P(A^c / B) = 4/7

6: Evaluate P(X∪Y), if 3P(X) = P(Y) = 2/5 and P(X/Y) = 4/5.

Given: 3P(X) = 2/5

⇒ P(X) = 2/15

and P(Y) = 2/5

Using Formula, P(X∩Y) = P(X / Y) × P(Y)

⇒ P(X∩Y) = 4/5 × 2/5

⇒ P(X∩Y) = 8/25

Now, P(X∪Y) = P(X) + P(Y) – P(X∩Y)

⇒ P(X∪Y) = 2/15 + 2/5 – 8/25

⇒ P(X∪Y) = 16/75

7: If P(A) = 5/8, P(B) = 1/4 and P(A /B) = 3 / 8. Find P(A^c/ B^c).

As we know, P(B^c) = 1 – P(B)

⇒ P(B^c) = 1 – 1/4

Thus, P(B^c) = 3/4

Using Fromula, P(A∩B) = P(A / B) × P(B)

⇒ P(A∩B) = 3/8 × 1/4

⇒ P(A∩B) = 3/32

Now, P(A∪B) = P(A) + P(B) – P(A∩B)

⇒ P(A∪B) = 5/8 + 1/4 -3/32

⇒ P(A∪B) =25/32

Again, P(A^c ∩ B^c) = 1 – P(A∪B)

⇒ P(A^c ∩ B^c)= 1 – 25/32

⇒ P(A^c ∩ B^c)= 7/32

⇒ P(A^c/ B^c) = P(A^c ∩ B^c) /P(B^c)

⇒ P(A^c/ B^c) = (7/32)/(3/4)

Thus, P(A^c/ B^c) = 7/24

8: In a school there are 100 students out of which 40 are boys. It is known that out of 40, 20% of boys study in class 11. What is the probability P(A/B) where A: student chosen randomly studies in class 11 and B: the chosen student is a boy.

n(A∩B) = 20% × 40 = 8

n(B) = 40

Using Formula, P(A / B) = n(A∩B) / n(B)

⇒ P(A / B) = 8 / 40

⇒ P(A / B) = 1/5

9: A coin is tossed 2 times. Find P(Y/X) if X = at least one head and Y = no tail.

Sample Space for 2 coin toss is:

S = {HH, HT, TH, TT}

X = {HH, HT, TH} and Y = {HH}

Thus, P(X) = 3/4,

P(Y) = 1/4, and

P(X∩Y) = 1/4

Thus, P(Y/X) = (1/4) / (3/4)

⇒ P(Y/X) = 1 / 3

10: A coin is tossed 3 times. Find (X/Y) if X = at most one head and Y = at most 2 tail.

Sample Space for 3 coin toss is:

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

P(X) = 4 / 8 = 1/2

P(Y) = 7/8

P(X∩Y) = 3/8

Now, P(X / Y) = P(X∩Y) / P(Y)

⇒ P(X / Y) = (3/8) / (7/8)

⇒ P(X / Y) = 3/7

Conditional Probability Practice Questions – Unsolved

Solve these worksheet containing Conditional Probability Practice Questions to test your understanding on the topic of Conditional probability.

1: If P(A) = 0.25, P(B) = 0.6 and P(A∩B) = 0.5. Find P(A/B).

2: If P(A) = 0.3, P(B/A) = 0.4 and P(A∪B) = 0.5 then find P(B).

3: A dice is rolled. If X = {3, 5, 6}, Y = {3, 4} and Z = {1, 3, 6} then, find P(X/Y), P(Y/Z), P(X/Z).

4: A coin is tossed 4 times. Find P(C/D) in C = Head on second Toss and D = Tail on third toss

5: Given that the two numbers appearing on throwing two dice are different. Find the probability of the event the sum of numbers appearing on dice is 8 and the number 4 appears once.

6: Evaluate P(X∪Y), if 2P(X) = P(Y) = 4/11 and P(X/Y) = 2/9.

7: If P(A) = 5/8, P(B) = 2/3 and P (A B) = 3 / 8. Find P (A^c/ B^c).

8: In a school there are 120 students out of which 60 are boys. It is known that out of 40, 10% of boys study in class 11. What is the probability that a student chosen randomly studies in class 11 , given that the chosen student is a boy.

9: A coin is tossed 5 times. Find (X/Y) if X = at least two heads and Y = at most one tail.

10: A coin is tossed 4 times. Find (X/Y) if X = at most one head and Y = at most 3 tails.

Also Read:

Practice Questions on Pythagoras Theorem

Practice Questions on Average

Conditional Probability- FAQs

What is Conditional Probability?

When two events are associated with a random experiment, the probability of the first event under the condition that the second event has already occurred and the probability of second event is not equal to zero is called the conditional probability.

What is the Conditional Probability Basic Formula?

The conditional probability basic formula is given by:

P(X / Y) = P(X∩Y) / P(Y)

How to Represent Conditional Probability?

The representation of conditional probability if A has already occurred before B is P(B /A).