Conditional Probability Questions with Solutions: 12-Question Practice Set

Conditional probability errors usually begin before the arithmetic. A solver keeps the original sample space, reverses the condition, or treats overlapping events as independent. Conditional probability problems involve notation and independence, restricted sample spaces, without-replacement draws and Bayes-style reverse conditioning. Choose an option and write the denominator before reading each explanation. That denominator habit matters more than memorising another formula. Before calculating, name the conditioned event, discard outcomes outside it, and check whether the numerator is a joint event. If the question reverses a direction such as P(E|H) to P(H|E), list every route to the evidence before using Bayes' rule. Build broader exam practice through GATE CS Exam Preparation.

Conditional probability formulas to use before the questions

Calibrate the denominator on a commute example. Of the students who reach one exam centre, 70% travel by metro and 30% by bus. Late arrival happens for 4% of the metro group and 15% of the bus group. For the late-arrival event L,

P(L) = 0.70(0.04) + 0.30(0.15) = 0.028 + 0.045 = 0.073.

Now reverse the direction:

P(bus | L) = 0.30(0.15) / 0.073 = 0.045 / 0.073 ≈ 0.616.

The evidence "arrived late" changes the denominator from every student to the late group alone. The Probability for GATE CS guide places this idea in the full concept sequence.

Conditional probability questions 1-3: certain events and independence

Question 1 (GATE 2003)

Let P(E) denote the probability of the event E. Given P(A) = 1, P(B) = 1/2, the values of P(A | B) and P(B | A) respectively are

• A. 1/4, 1/2

• B. 1/2, 1/14

• C. 1/2, 1

• D. 1, 1/2

Answer: D.

Since P(A)=1, A∩B=B. Hence P(A|B)=P(B)/P(B)=1 and P(B|A)=P(B)/1=1/2. Solution.

Question 2 (GATE 1999)

Consider two events E₁ and E₂ such that P(E₁) = 1/2, P(E₂) = 1/3, and P(E₁ ∩ E₂) = 1/5. Which of the following statements is true?

• A. P(E₁ ∩ E₂) = 2/3

• B. Events E₁ and E₂ are independent.

• C. Events E₁ and E₂ are not independent.

• D. P(E₁ | E₂) = 4/5

Answer: C.

Independence needs (1/2)(1/3)=1/6, but the intersection is 1/5. Also, P(E₁|E₂)=(1/5)/(1/3)=3/5, ruling out D. Solution.

Question 3 (GATE 2023)

Consider a random experiment where two fair coins are tossed. Let A be the event that denotes HEAD on both the throws, B be the event that denotes HEAD on the first throw, and C be the event that denotes HEAD on the second throw. Which of the following statements is/are TRUE?

• A. A and B are independent.

• B. A and C are independent.

• C. B and C are independent.

• D. Prob(B|C) = Prob(B)

Answer: C and D.

On {HH, HT, TH, TT}, P(A)=1/4 and P(B)=P(C)=1/2. The A-B and A-C intersections are 1/4, not 1/8. But P(B∩C)=1/4=P(B)P(C) and P(B|C)=1/2=P(B). Solution.

Conditional probability questions 4-7: restrict the sample space first

Question 4 (Infosys 2023)

A black die and a red die are rolled. Find the conditional probability of obtaining a sum of 8, given that the red die shows a number less than 4.

• A. 1/9

• B. 4/9

• C. 5/9

• D. 7/9

Answer: A.

Red is restricted to 1, 2, 3, giving 3×6=18 outcomes. Two, (2,6) and (3,5) in red-black order, sum to 8. Thus 2/18=1/9. Solution.

Question 5 (DSSSB 2021)

20 percent of the pens produced in a factory are of red colour and 4 percent are red and defective. If one pen is picked up at random, what is the probability of its being defective if it is red?

• A. 0.2

• B. 0.4

• C. 0.6

• D. 0.8

Answer: A.

Red pens form the denominator: P(defective|red)=0.04/0.20=0.20. Using 0.04 alone would answer the joint event, not the conditional question. Solution.

Question 6 (TPSC 2025)

In a group of 100 students, 60 play cricket, 50 play football, and 30 play both. If a student is selected at random, what is the probability that the student plays cricket given that the student plays football ?

• A. 0.15

• B. 0.30

• C. 0.60

• D. 0.18

Answer: C.

Given football, the pool has 50 students, with 30 cricket players. Thus P(cricket|football)=30/50=0.60. Solution.

Question 7

A box contains 4 white balloons and 3 red balloons. In succession, two balloons are selected at random and removed from the box. Given that the first balloon removed is white, what is the probability that the second balloon removed is red?

• A. 4/7

• B. 3/7

• C. 1/2

• D. 12/42

Answer: C.

After one white is removed, 3 white and 3 red remain. Thus P(second red|first white)=3/6=1/2. Option B keeps the denominator at 7, which still counts the balloon that was already removed. Continue the same no-replacement pattern in the Conditional Probability PYQ Questions hub.

Conditional probability questions 8-10: Bayes and reverse conditioning

Question 8 (BEL 2007)

Company X shipped 5 computer chips, 1 of which was defective, and Company Y shipped 4 computer chips, 2 of which were defective. One computer chip is to be chosen uniformly at random from the 9 chips shipped by the companies. If the chosen chip is found to be defective, what is the probability that the chip came from Company Y ?

• A. 2/9

• B. 4/9

• C. 2/3

• D. 1/2

Answer: C.

The defective pool has one X chip and two Y chips. Hence P(Y|defective)=2/3. Solution.

Question 9 (TPSC 2024)

A multiple choice examination has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices.

As a teacher you want the test to measure what the student knows. If the student answers a question correctly, what is the probability that they knew the answer ?

• A. 0.61

• B. 0.67

• C. 0.71

• D. 0.77

Answer: D.

P(correct)=1/2+1/12+1/16=31/48. Since P(know∩correct)=1/2, Bayes gives (1/2)/(31/48)=24/31≈0.774, closest to D. Solution.

Question 10 (UGC NET 2022)

There are three boxes. First box has 2 white, 3 black and 4 red balls. Second box has 3 white, 2 black and 2 red balls. Third box has 4 white, 1 black and 3 red balls. A box is chosen at random and 2 balls are drawn out of which 1 is white, and 1 is red. What is the probability that the balls came from first box?

• A. 0.237

• B. 0.723

• C. 0.18

• D. 0.452

Answer: A.

The likelihoods are L₁=2×4/C(9,2)=2/9, L₂=3×2/C(7,2)=2/7 and L₃=4×3/C(8,2)=3/7. Equal priors cancel, so P(box1|WR)=(2/9)/(2/9+2/7+3/7)=14/59≈0.237. Solution.

Conditional probability questions 11-12: reconstruct the missing probabilities

Question 11 (GATE 2009)

An unbalanced die (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3?

• A. 0.453

• B. 0.468

• C. 0.485

• D. 0.492

Answer: B.

Let each even face have probability e. Then 3e+2.7e=1, so e=10/57. Faces 4 and 6 give 20/57. Because (20/57)/P(>3)=0.75, P(>3)=80/171≈0.468. Solution.

Question 12 (GATE 2017)

P and Q are considering to apply for a job. The probability that P applies for the job is 1/4, the probability that P applies for the job given that Q applies for the job is 1/2, and the probability that Q applies for the job given that P applies for the job is 1/3. Then the probability that P does not apply for the job given that Q does not apply for this job is

• A. 4/5

• B. 5/6

• C. 7/8

• D. 11/12

Answer: A.

P(P∩Q)=(1/3)(1/4)=1/12, then P(P|Q)=1/2 gives P(Q)=1/6. So P(P^c∩Q^c)=1-(1/4+1/6-1/12)=2/3 and P(Q^c)=5/6. Their ratio is (2/3)/(5/6)=4/5. Solution.

Common conditional probability traps and their repairs

In a timed solve, write the condition right of the bar, circle the denominator, and calculate the joint numerator. Reject answers above 1 or against an obvious count. For Bayes, sketch a two- or three-branch tree. Repair weak conditioned counting with Permutations and Combinations for GATE CS.

Conditional probability practice: the next step

Retest yourself on Questions 3, 4, 7, 9, 10 and 12 without looking at the answers. Questions 3, 4, 7, 9, 10 and 12 use independence, restricted outcomes, draws without replacement, total probability, Bayes' rule and complement events. Write one denominator line for every answer, even when the count seems obvious. Then compare the route, not just the option. Once you know which step fails, use Engineering Mathematics for GATE as a sequenced route through the wider mathematics syllabus.

The short version is simple: the event after the bar defines the universe. Bayes reverses direction only after every evidence path is counted. Independence must be tested, never assumed.