Conditional Probability Practice Question

Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. In simple words, conditional probability is like figuring out the chances of something happening given that something else has already happened.

The probability of A given that B has already occurred is denoted by : P(A|B)

Imagine you're in a candy store that has 100 candies: 30 are chocolate, and 70 are fruit candies. If you know you picked a candy but didn't see what it was, the chance of it being chocolate is 30 out of 100, or 30%.

Now, suppose someone tells you that the candy you picked is in a blue wrapper, and only chocolate candies have blue wrappers. Given this new information, you now know for sure that the candy you picked is chocolate.

Important Formulas on Conditional Probability

The below table represents the different conditional probability formulas.

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.

Question 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).

Solution:

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

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

Solution:

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

Question 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).
Solution:

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

Question 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

Solution:

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

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

Solution:

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

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

Solution:

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

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

Solution:

As we know, P(B^c) = 1 - P(B)
⇒ P(B^c) = 1 - 1/4
Thus, P(B^c) = 3/4

Using Formula, 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

Question 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.

Solution:

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

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

Solution:

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

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

Solution:

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

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

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

Question 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).

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

Question 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.

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

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

Question 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.

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

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

Also Read:

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• Practice Questions on Average