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    Conditional Probability and Bayes Theorem PYQs for GATE CS

    Solve 5+ Conditional Probability and Bayes Theorem previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

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    Question 1
    2026 Slot Set2 PYQ
    Level 3: Exam Standard
    Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.

    The conditional probability is equal to ____________. (rounded off to one decimal place)
    Question 2
    2025 Slot Set1 PYQ
    Level 3: Exam Standard

    A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability and for a fake coin, . You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)

    Question 3
    2024 Slot Set1 PYQ
    Level 3: Exam Standard

    A bag contains 10 red balls and 15 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (<i>rounded off to 3 decimal places</i>) that both balls drawn are red is _________

    Question 4
    2022 PYQ
    Level 3: Exam Standard
    A box contains five balls of same size and shape. Three of them are green coloured balls and two of them are orange coloured balls. Balls are drawn from the box one at a time. If a green ball is drawn, it is not replaced. If an orange ball is drawn, it is replaced with another orange ball.

    First ball is drawn. What is the probability of getting an orange ball in the next draw?
    Question 5
    2021 Slot Set1 PYQ
    Level 3: Exam Standard
    A sender (S) transmits a signal, which can be one of the two kinds: and with probabilities 0.1 and 0.9 respectively, to a receiver (R).
    In the graph below, the weight of edge is the probability of receiving when is transmitted, where . For example, the probability that the received signal is given the transmitted signal was , is 0.7.

    S R H L H L 0.3 0.7 0.8 0.2
    If the received signal is , the probability that the transmitted signal was (rounded to 2 decimal places) is __________.
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    Conditional Probability and Bayes Theorem PYQs for GATE CS

    Solve 5+ Conditional Probability and Bayes Theorem previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

    Chapter Roadmap: Conditional Probability and Bayes Theorem

    Chapter Roadmap

    By the end of this chapter, you will master the art of updating probabilities with new evidence, solving multi-stage sampling problems, and analyzing repeated independent trials.

    Step 1 (Current)
    Bayes Theorem and Posterior Inference
    Reversing conditional probability. Updating prior beliefs with new evidence to find posterior probabilities.
    Step 2
    Conditional Probability in Sequential Sampling
    Drawing items with or without replacement. Tree diagrams and path probabilities.
    Step 3
    Conditional Events in Repeated Coin Tosses
    Analyzing specific patterns in sequences of independent Bernoulli trials.

    Bayes Theorem and Posterior Inference

    Bayes Theorem and Posterior Inference

    The mathematics of updating beliefs with new evidence.

    What you will master here:
    • Reversing conditional probability from effect to cause
    • Identifying Prior, Likelihood, Marginal, and Posterior components
    • Applying the Law of Total Probability to compute denominators
    • Solving classic coin, medical testing, and communication channel problems

    Conditional Probability and Bayes Theorem: Solved Questions with Step-by-Step Explanations (5 Problems)

    Question 1 · Engineering Mathematics · 2026_Set2 NAT
    Suppose an unbiased coin is tossed 6 times. Each coin toss is independent of all previous coin tosses. Let be the event that among the second, fourth, and sixth coin tosses, there are at least two heads. Let be the event that among the first, second, third, and fifth coin tosses, there are equal number of heads and tails.

    The conditional probability is equal to ____________. (rounded off to one decimal place)
    Correct Answer:

    0.5

    Step-by-Step Solution

    Insight: This is a conditional probability problem with overlapping events in repeated trials. The shared variable (Toss 2) prevents independence.

    Exam route: Use the reduced sample space method. Count outcomes for E2 (denominator), then count outcomes where both E1 and E2 hold by splitting on the shared Toss 2 (numerator).

    Learning route:

    Step 1: Understand the events.

    • E1: Tosses {2, 4, 6} have >= 2 Heads.
    • E2: Tosses {1, 2, 3, 5} have exactly 2 Heads and 2 Tails.

    Step 2: Calculate the denominator N(E2).

    • Tosses {1, 2, 3, 5} must have 2H, 2T. Number of ways = C(4, 2) = 6.
    • Tosses {4, 6} are unconstrained by E2. Number of ways = 2^2 = 4.
    • Total N(E2) = 6 * 4 = 24.

    Step 3: Calculate the numerator N(E1 ∩ E2) by conditioning on the shared Toss 2.

    Case A: Toss 2 is Head (H).

    • For E2 to hold, the remaining tosses in its set {1, 3, 5} must have exactly 1H, 2T. Ways = C(3, 1) = 3.
    • For E1 to hold, the remaining tosses in its set {4, 6} must have >= 1H (since Toss 2 is already H). Ways = Total - 0H = 4 - 1 = 3.
    • Ways for Case A = 3 * 3 = 9.

    Case B: Toss 2 is Tail (T).

    • For E2 to hold, {1, 3, 5} must have exactly 2H, 1T. Ways = C(3, 2) = 3.
    • For E1 to hold, {4, 6} must have >= 2H (since Toss 2 is T). Ways = 1 (both must be H).
    • Ways for Case B = 3 * 1 = 3.
    • Total N(E1 ∩ E2) = 9 + 3 = 12.

    Step 4: Compute the conditional probability.

    P(E1 | E2) = N(E1 ∩ E2) / N(E2) = 12 / 24 = 0.5.

    Verification: Plug back into the definition. P(E1 ∩ E2) = 12/64, P(E2) = 24/64. Ratio is 12/24 = 0.5. Matches perfectly.

    Question 2 · Engineering Mathematics · 2025_Set1 NAT

    A box contains 5 coins: 4 regular coins and 1 fake coin. When a regular coin is tossed, the probability and for a fake coin, . You pick a coin at random and toss it twice, and get two heads. The probability that the coin you have chosen is the fake coin is _______. (rounded off to two decimal places)

    Correct Answer:

    0.50

    Step-by-Step Solution

    Insight: This is a classic Bayes Theorem posterior inference problem, reversing the conditional from "two heads given coin type" to "coin type given two heads".

    Exam route: Define priors P(Fake)=0.2, P(Regular)=0.8. Likelihoods P(2H|Fake)=1, P(2H|Regular)=0.25. Apply Bayes formula directly.

    Learning route:

    Step 1: Define events and priors.

    • F: Chosen coin is fake. P(F) = 1/5 = 0.2.
    • R: Chosen coin is regular. P(R) = 4/5 = 0.8.
    • E: Observed two heads in two tosses.

    Step 2: Determine likelihoods.

    • P(E | F) = 1 * 1 = 1.0.
    • P(E | R) = 0.5 * 0.5 = 0.25.

    Step 3: Compute the denominator (Total Probability of E).

    • P(E) = P(E | F)P(F) + P(E | R)P(R)
    • P(E) = (1.0 0.2) + (0.25 0.8) = 0.2 + 0.2 = 0.4.

    Step 4: Apply Bayes Theorem for the posterior.

    • P(F | E) = P(E | F)P(F) / P(E) = 0.2 / 0.4 = 0.5.

    Verification: The posterior probabilities must sum to 1. P(R | E) = (0.25 * 0.8) / 0.4 = 0.2 / 0.4 = 0.5. Sum = 0.5 + 0.5 = 1.0. The calculation is sound.

    Question 3 · Engineering Mathematics · 2024_Set1 NAT

    A bag contains 10 red balls and 15 blue balls. Two balls are drawn randomly without replacement. Given that the first ball drawn is red, the probability (<i>rounded off to 3 decimal places</i>) that both balls drawn are red is _________

    Correct Answer:

    0.375

    Step-by-Step Solution

    Insight: This is a "Given the First Draw" sequential sampling problem. The condition explicitly updates the state of the bag, simplifying the problem to a single conditional probability.

    Exam route: Since the first ball is known to be red, update the bag's contents (9 Red, 15 Blue, Total 24) and directly compute the probability of drawing a red ball from this new state.

    Learning route:

    Step 1: Identify the initial state.

    • 10 Red, 15 Blue. Total = 25.

    Step 2: Apply the given condition to update the state.

    • Condition: The first ball drawn is Red.
    • Action: Remove 1 Red ball (without replacement).
    • New State: 9 Red, 15 Blue. Total = 24.

    Step 3: Calculate the probability of the target event from the new state.

    • Target: The second ball is Red (which satisfies "both balls drawn are red" given the first is red).
    • P(2nd Red | 1st Red) = (Number of remaining Red) / (New Total) = 9 / 24.

    Step 4: Simplify and format the answer.

    • 9 / 24 = 3 / 8 = 0.375.

    Verification: The probability of the complement (2nd is Blue | 1st is Red) is 15 / 24 = 5 / 8 = 0.625. Sum = 0.375 + 0.625 = 1.0. The state update is correct.

    Question 4 · Engineering Mathematics · 2022 MCQ
    A box contains five balls of same size and shape. Three of them are green coloured balls and two of them are orange coloured balls. Balls are drawn from the box one at a time. If a green ball is drawn, it is not replaced. If an orange ball is drawn, it is replaced with another orange ball.

    First ball is drawn. What is the probability of getting an orange ball in the next draw?
    1. A.

    2. B.

    3. C.

    4. D.

    Correct Answer:

    D

    Step-by-Step Solution

    Insight: This is an asymmetric replacement problem requiring the Law of Total Probability over the unknown first draw.

    Exam route: Branch into two paths: 1st is Green (prob 3/5, new state 2G, 2O) and 1st is Orange (prob 2/5, new state 3G, 2O). Multiply and sum the path probabilities.

    Learning route:

    Step 1: Identify initial state: 3 Green (G), 2 Orange (O). Total = 5.

    Step 2: Define the two mutually exclusive paths for the first draw.

    Path A: First ball is Green.

    • Probability of Path A: P(1st G) = 3/5.
    • Rule: Green is not replaced. New state: 2 G, 2 O. Total = 4.
    • Probability of 2nd Orange given Path A: P(2nd O | 1st G) = 2/4 = 1/2.
    • Joint probability of Path A: (3/5) * (1/2) = 3/10 = 15/50.

    Path B: First ball is Orange.

    • Probability of Path B: P(1st O) = 2/5.
    • Rule: Orange is replaced with another Orange. The drawn orange is removed, but another is added, so the count of Orange remains 2, and total remains 5. New state: 3 G, 2 O. Total = 5.
    • Probability of 2nd Orange given Path B: P(2nd O | 1st O) = 2/5.
    • Joint probability of Path B: (2/5) * (2/5) = 4/25 = 8/50.

    Step 3: Apply the Law of Total Probability.

    P(2nd O) = P(Path A) + P(Path B) = 15/50 + 8/50 = 23/50.

    Verification: The sum of all path probabilities for the second draw must equal 1. P(2nd G) = (3/5 2/4) + (2/5 3/5) = 15/50 + 12/50 = 27/50. Total = 23/50 + 27/50 = 1. The math is perfectly consistent.

    Question 5 · Engineering Mathematics · 2021_Set1 NAT
    A sender (S) transmits a signal, which can be one of the two kinds: and with probabilities 0.1 and 0.9 respectively, to a receiver (R).
    In the graph below, the weight of edge is the probability of receiving when is transmitted, where . For example, the probability that the received signal is given the transmitted signal was , is 0.7.

    S R H L H L 0.3 0.7 0.8 0.2
    If the received signal is , the probability that the transmitted signal was (rounded to 2 decimal places) is __________.
    Correct Answer:

    0.04

    Step-by-Step Solution

    Insight: This is a communication channel problem, which is a direct application of Bayes Theorem where edge weights are likelihoods and node frequencies are priors.

    Exam route: Identify P(T=H)=0.1, P(T=L)=0.9. Likelihoods from graph: P(R=H|T=H)=0.3, P(R=H|T=L)=0.8. Apply Bayes formula for P(T=H|R=H).

    Learning route:

    Step 1: Extract priors from the problem statement.

    • P(T=H) = 0.1
    • P(T=L) = 0.9

    Step 2: Extract likelihoods from the graph edge weights.

    • P(R=H | T=H) = 0.3
    • P(R=H | T=L) = 0.8

    Step 3: Compute the marginal probability of receiving H (the denominator).

    • P(R=H) = P(R=H | T=H)P(T=H) + P(R=H | T=L)P(T=L)
    • P(R=H) = (0.3 0.1) + (0.8 0.9) = 0.03 + 0.72 = 0.75.

    Step 4: Apply Bayes Theorem to find the posterior.

    • P(T=H | R=H) = P(R=H | T=H)P(T=H) / P(R=H)
    • P(T=H | R=H) = 0.03 / 0.75 = 3 / 75 = 1 / 25 = 0.04.

    Verification: Calculate P(T=L | R=H) = 0.72 / 0.75 = 0.96. Sum of posteriors = 0.04 + 0.96 = 1.0. The result is consistent.

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