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.

Question 1: A bag contains 5 red balls and 7 blue balls. Two balls are drawn without replacement. What is the probability that the second ball drawn is red, given that the first ball drawn was red?
Solution:

Let the events be,

Event A: The first ball drawn is red.
Event B: The second ball drawn is red.

P(A) = 5/12 and P(B) = 4/11 (as first ball drawn is already red, thus only 4 red balls remain in the bag)

Therefore, probability of the second ball drawn being red given that the first ball drawn was red is 4/11.

Question 4: In a bag, there are 4 red balls and 6 blue balls. Two balls are drawn without replacement. What is the probability that the second ball is red, given that the first ball drawn was blue?
Solution:

After drawing a blue ball, there are 4 red balls and 5 blue balls left.
P(Second is red | First was blue) = 4 / (4 + 5) = 4/9 = 0.444

Question 2: In a survey among a group of students, 70% play football, 60% play basketball, and 40% play both sports. If a student is chosen at random and it is known that the student plays basketball, what is the probability that the student also plays football?
Solution:

Let's assume there are 100 students in the survey.

Number of students who play football = n(A) = 70
Number of students who play basketball = n(B) = 60
Number of students who play both sports = n(A ∩ B) = 40

To find the probability that a student plays football given that they play basketball, we use the conditional probability formula:
P(A|B) = n(A ∩ B) / n(B)

Substituting the values, we get:
P(A|B) = 40 / 60 = 2/3

Therefore, probability that a randomly chosen student who plays basketball also plays football is 2/3.

Question 3: A fair die is rolled twice. Given that the sum of the two rolls is even, what is the probability that the first roll was an even number?
Solution:

Total favorable outcomes for an even sum:

Both rolls even: 9 outcomes (e.g., (2, 2), (2, 4), ...)

Both rolls odd: 9 outcomes (e.g., (1, 1), (1, 3), ...)

Total = 18 outcomes.

Favorable outcomes where the first roll is even:

First roll even (2, 4, 6), second roll even (2, 4, 6): 9 outcomes.

Conditional probability: P(First roll even | Sum even) = 9 / 18 = 1 / 2

Question 4: 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: 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 11: 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 12: 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 13: 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) = 0.15. 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 or 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 60, 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.