chapter

Conditional Probability and Bayes' Theorem PYQs for GATE DA

Solve 4+ Conditional Probability and Bayes' Theorem previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Try a question

Answer it here to see how it works. Nothing is recorded until you sign in.

Question 1

2026 PYQ

Level 3: Exam Standard

A clinic specializes in testing for a disease D. The result of the test can be either positive or negative. A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals. If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places)

Question 2

2025 PYQ

Level 3: Exam Standard

There are three boxes containing white balls and black balls. Box-1 contains 2 black and 1 white balls. Box-2 contains 1 black and 2 white balls. Box-3 contains 3 black and 3 white balls. In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is \frac{1}{2}, Box-2 is \frac{1}{6}, and Box-3 is \frac{1}{3} . A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is (Round off to two decimal places)

Question 3

2025 PYQ

Level 3: Exam Standard

The naive Bayes classifier is used to solve a two-class classification problem with class-labels y_{1}, y_{2} . Suppose the prior probabilities are P (y_{1}) = \frac{1}{3} and P (y_{2}) = \frac{2}{3} . Assuming a discrete feature space with P (x ∣ y_{1}) = \frac{3}{4} and P (x ∣ y_{2}) = \frac{1}{4} for a specific feature vector x . The probability of misclassifying x is (Round off to two decimal places)

Question 4

2024 PYQ

Level 3: Exam Standard

Consider two events T and S. Let \overline{T} denote the complement of the event T. The probability associated with different events are given as follows: P (\overline{T}) = 0.6, P (S ∣ T) = 0.3, P (S ∣ \overline{T}) = 0.6 Then, P (T ∣ S) is ______ (rounded off to two decimal places).

Free preview ends here

Login to view the complete previous-year questions and solutions

Creating an account is free. You get the rest of this chapter, step-by-step solutions, and a study plan built around the topics you are actually weak at.

Why MastersUp

Personalised first. High quality throughout.

Most platforms hand everyone the same content. Here the content moves with your performance, topic by topic.

Built around you, not around a syllabus PDF

Every answer you give moves your topic-level intelligence rate. The next question, the next revision card and tomorrow's plan all change with it.

Revision that hits your weak spots

We only revise topics you have actually attempted and are still below the safe bar on — never the same chapter on repeat.

Questions calibrated to the real exam

Each question carries a measured toughness. You are served a rung above your current level, so practice keeps stretching you.

Notes written for recall, not for volume

Full lesson cards for first study, curated short-note cards for the last mile — with derivations, traps and exam patterns marked.

One place for everything

Notes, chapter practice, previous-year questions, test series and full-length papers — all feeding one picture of your preparation.

Honest progress

No vanity streaks. Progress here means chapters mastered and accuracy that held up on harder questions.

Unlock the whole course

Full notes and short notes, the complete question bank with worked solutions, mock tests, full-length papers, and an adaptive plan that rebuilds itself as you improve.

View all plans Start free

Conditional Probability and Bayes' Theorem PYQs for GATE DA

Solve 4+ Conditional Probability and Bayes' Theorem previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Conditional Probability and Bayes

Probability · Unit 1 · Chapter 2

Conditional Probability & Bayes' Theorem

Reverse the direction of inference. Update beliefs with evidence. Classify with posteriors.

Core exam weapon

Stage 01 · Foundation

Conditional Probability & Multiplication Rule

Reshape the sample space when information arrives.

Stage 02 · Decomposition

Law of Total Probability

Break a hard event into a sum over cases.

Stage 03 · Reversal

Bayes' Theorem

Infer causes from observed effects.

Stage 04 · Application

Posterior Classification & Naive Bayes

From theorem to classifier, with misclassification risk.

Conditional Probability, Bayes and Posterior Classification

Topic 1 of 1 · Core

Conditional Probability & Bayes' Theorem

Condition on evidence. Reverse the inference. Classify with posteriors.

Conditioning intuition

Total probability

Bayes' theorem

Naive Bayes classifier

Conditional Probability and Bayes' Theorem: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Probability and Statistics · 2026 NAT

A clinic specializes in testing for a disease D. The result of the test can be either positive or negative.

A study revealed that if a person suffers from the disease D, the test result in that clinic comes out positive 80% of the time, and negative 20% of the time. If a person is not suffering from the disease D, the test comes out positive 10% of the time and negative 90% of the time. It is also known that among the general population, the disease D occurs in 30% of the individuals.

If a person tests positive for D in that clinic, the probability that he/she actually suffers from the disease D is __________ . (Rounded off to two decimal places)

Correct Answer:

0.77

Step-by-Step Solution

Insight: This is a textbook Bayes-reversal question — you know $P (+ ∣ D)$ and $P (D)$ , and you want $P (D ∣ +)$ . The denominator is the law of total probability over ${D, \overline{D}}$ .

Exam route:

Step 1 — Write priors and likelihoods:

$P (D) = 0.3, P (\overline{D}) = 0.7$

$P (+ ∣ D) = 0.8, P (+ ∣ \overline{D}) = 0.1$

Step 2 — Total probability of a positive test:

$P (+) = P (+ ∣ D) P (D) + P (+ ∣ \overline{D}) P (\overline{D}) = 0.8 (0.3) + 0.1 (0.7) = 0.24 + 0.07 = 0.31$

Step 3 — Bayes:

$P (D ∣ +) = \frac{P ( + ∣ D ) P ( D )}{P ( + )} = \frac{0.24}{0.31} = \frac{24}{31} \approx 0.7742$

Rounded to two decimal places: $0.77$ .

Learning route: The event "test positive" can happen in two mutually exclusive ways — the person has the disease and the test correctly flags it, or the person is healthy and the test falsely flags it. These two branches cover every way to see $+$ . Sum their joint probabilities to get $P (+)$ , then the diseased branch divided by the total gives the posterior.

Trap check: If you swap numerator and denominator and compute $P (+ ∣ D)$ instead of $P (D ∣ +)$ , you get $0.80$ — a classic reversal error.

Verification: $P (D ∣ +) + P (\overline{D} ∣ +) = 24/31 + 7/31 = 1$ . ✓

Question 2 · Probability and Statistics · 2025 NAT

There are three boxes containing white balls and black balls.

Box-1 contains 2 black and 1 white balls.
Box-2 contains 1 black and 2 white balls.
Box-3 contains 3 black and 3 white balls.

In a random experiment, one of these boxes is selected, where the probability of choosing Box-1 is

\frac{1}{2}

, Box-2 is

\frac{1}{6}

, and Box-3 is

\frac{1}{3}

. A ball is drawn at random from the selected box. Given that the ball drawn is white, the probability that it is drawn from Box-2 is
(Round off to two decimal places)

Correct Answer:

0.25

Step-by-Step Solution

Insight: Three-box Bayes. The partition is ${B_{1}, B_{2}, B_{3}}$ , the evidence is "white ball drawn". You want $P (B_{2} ∣ W)$ .

Exam route:

Step 1 — Priors and likelihoods:

$P (B_{1}) = \frac{1}{2}, P (B_{2}) = \frac{1}{6}, P (B_{3}) = \frac{1}{3}$

$P (W ∣ B_{1}) = \frac{1}{3}, P (W ∣ B_{2}) = \frac{2}{3}, P (W ∣ B_{3}) = \frac{1}{2}$

Step 2 — Branches:

$P (W ∣ B_{1}) P (B_{1}) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$

$P (W ∣ B_{2}) P (B_{2}) = \frac{2}{3} \cdot \frac{1}{6} = \frac{2}{18} = \frac{1}{9}$

$P (W ∣ B_{3}) P (B_{3}) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$

Step 3 — Total:

$P (W) = \frac{1}{6} + \frac{1}{9} + \frac{1}{6} = \frac{3 + 2 + 3}{18} = \frac{8}{18} = \frac{4}{9}$

Step 4 — Posterior:

$P (B_{2} ∣ W) = \frac{1/9}{4/9} = \frac{1}{4} = 0.25$

Learning route: Each box is a hypothesis. The evidence "white ball" is more likely under Box 2 than Box 1 or 3, but Box 2 is chosen less often a priori. Bayes balances these two forces.

Trap check: If you forget that Box 3 has 3 white out of 6 and write $P (W ∣ B_{3}) = 3$ or $1/3$ , the denominator breaks. Always reduce the fraction first.

Verification: $P (B_{1} ∣ W) + P (B_{2} ∣ W) + P (B_{3} ∣ W) = \frac{3/18}{8/18} + \frac{2/18}{8/18} + \frac{3/18}{8/18} = \frac{3 + 2 + 3}{8} = 1$ . ✓

Question 3 · Probability and Statistics · 2025 NAT

The naive Bayes classifier is used to solve a two-class classification problem with class-labels

y_{1}, y_{2}

. Suppose the prior probabilities are

P (y_{1}) = \frac{1}{3}

and

P (y_{2}) = \frac{2}{3}

. Assuming a discrete feature space with

P (x ∣ y_{1}) = \frac{3}{4}

and

P (x ∣ y_{2}) = \frac{1}{4}

for a specific feature vector

x

. The probability of misclassifying

x

is
(Round off to two decimal places)

Correct Answer:

0.40

Step-by-Step Solution

Insight: This is a MAP-decision misclassification question. Compute both posteriors, pick the larger, and the misclassification probability is the posterior of the other class. Exam route: Step 1 — Unnormalised posteriors:

P (x ∣ y_{1}) P (y_{1}) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}

P (x ∣ y_{2}) P (y_{2}) = \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{6}

Step 2 — Normalise: Sum

= \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}

P (y_{1} ∣ x) = \frac{1/4}{5/12} = \frac{3}{5} = 0.6

P (y_{2} ∣ x) = \frac{1/6}{5/12} = \frac{2}{5} = 0.4

Step 3 — MAP rule: predict

y_{1}

(since

0.6 > 0.4

). Misclassification

= P (y_{2} ∣ x) = 0.40

. Learning route: A Bayes classifier assigns

x

ar g max_{i} P (y_{i} ∣ x)

. The probability of being wrong is exactly the posterior of the class not chosen. Because the denominator is common, you can compare the unnormalised scores directly to find the winner, then normalise only to read off the loser's posterior. Trap check: If you misclassify as

y_{2}

you would report

0.60

— but MAP picks the larger posterior, so

y_{1}

is the prediction and

y_{2}

is the error. Verification:

P (y_{1} ∣ x) + P (y_{2} ∣ x) = 0.6 + 0.4 = 1

. ✓

Question 4 · Probability and Statistics · 2024 NAT

Consider two events T and S. Let

\overline{T}

denote the complement of the event T. The
probability associated with different events are given as follows:

P (\overline{T}) = 0.6, P (S ∣ T) = 0.3, P (S ∣ \overline{T}) = 0.6

Then,

P (T ∣ S)

is ______ (rounded off to two decimal places).

Correct Answer:

0.25

Step-by-Step Solution

Insight: You are given $P (\overline{T})$ and two conditionals $P (S ∣ T), P (S ∣ \overline{T})$ , and asked for $P (T ∣ S)$ . First recover $P (T)$ , then use Bayes.

Exam route:

Step 1 — Recover the prior:

$P (\overline{T}) = 0.6 \Rightarrow P (T) = 1 - 0.6 = 0.4$

Step 2 — Total probability of $S$ :

$P (S) = P (S ∣ T) P (T) + P (S ∣ \overline{T}) P (\overline{T}) = 0.3 (0.4) + 0.6 (0.6) = 0.12 + 0.36 = 0.48$

Step 3 — Bayes:

$P (T ∣ S) = \frac{P ( S ∣ T ) P ( T )}{P ( S )} = \frac{0.12}{0.48} = \frac{1}{4} = 0.25$

Learning route: The complement $P (\overline{T})$ is a disguised prior — flip it to get $P (T)$ first. Then the problem is a standard two-branch Bayes with partition ${T, \overline{T}}$ and evidence $S$ .

Trap check: If you plug $P (\overline{T}) = 0.6$ directly into the numerator as if it were $P (T)$ , you get $0.3 (0.6) / P (S) = 0.18/0.54 = 1/3 \approx 0.33$ — a classic complement-mix-up.

Verification: $P (T ∣ S) + P (\overline{T} ∣ S) = 0.12/0.48 + 0.36/0.48 = 0.25 + 0.75 = 1$ . ✓

Try a question

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Conditional Probability and Bayes' Theorem PYQs for GATE DA

Chapter Roadmap: Conditional Probability and Bayes

Conditional Probability & Bayes' Theorem

Conditional Probability, Bayes and Posterior Classification

Conditional Probability & Bayes' Theorem

Conditional Probability and Bayes' Theorem: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit