Urn Models and Reinforcement Processes PYQs for GATE CS

Solve 2+ Urn Models and Reinforcement Processes previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Urn Models and Reinforcement Processes

Your Journey Through Urn Models

Current Focus: Polya Urn Reinforcement Processes

The reinforcement mechanism and state evolution
Exchangeability and why order does not matter
Computing probabilities at any step
Connection to Beta-Binomial distributions
Martingale properties and convergence

Why This Matters for GATE

Polya Urn models test your understanding of conditional probability in sequential processes, the difference between independent and dependent events, and the long-term behavior of stochastic systems. Master the pattern, and you will recognize it instantly.

The Polya Urn: Rich Get Richer

Start with $r$ red and $b$ black balls. At each step, draw one ball, note its color, and return it plus one additional ball of the same color.

If Red Drawn	If Black Drawn
Red: $r \to r + 1$	Black: $b \to b + 1$
Total: $r + b + 1$	Total: $r + b + 1$

Key Insight: Early outcomes get amplified. This is not independent trials — each draw changes the composition of the urn.

Urn Models and Reinforcement Processes: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Engineering Mathematics · 2026_Set1 MCQ

An urn contains one red ball and one blue ball. At each step, a ball is picked uniformly at random from the urn, and this ball together with another ball of the same color is put back in the urn. The probability that there are equal number of red and blue balls after two steps is

A.
$\frac{1}{4}$
B.
$\frac{1}{3}$
C.
$\frac{1}{2}$
D.
$\frac{2}{3}$

Correct Answer:

Step-by-Step Solution

Insight: Equal red and blue after 2 steps means exactly 1 red and 1 blue were drawn — use exchangeability or direct enumeration.

Exam route: Start: 1R, 1B. After 2 steps, total = 4. Equal counts need 2R and 2B, so exactly $k = 1$ red drawn. By exchangeability: $P = (1 2) \cdot 1^{(1)} \cdot 1^{(1)} / 2^{(2)} = 2 \cdot 1 \cdot 1/6 = 1/3$ . Select B.

Learning route:

This is a Polya Urn state question, identifiable by the reinforcement mechanism (draw, replace, add one of the same colour) and the question asking about the urn's composition after a fixed number of steps.

Setup: Initial state: 1 red, 1 blue, total 2. After 2 draws, total = $2 + 2 = 4$ balls. For equal red and blue counts, we need 2 red and 2 blue, which means exactly 1 red and 1 blue were drawn in the 2 draws.

Method 1 — Direct enumeration:

List all possible sequences of 2 draws:

RR: $P = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ → final: 3R, 1B (not equal)
RB: $P = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ → final: 2R, 2B (equal ✓)
BR: $P = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$ → final: 2R, 2B (equal ✓)
BB: $P = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$ → final: 1R, 3B (not equal)

$P (equal) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

Method 2 — Exchangeability formula:

$P (k reds in n draws) = \frac{( k n ) \cdot r ^{(k)} \cdot b ^{(n - k)}}{( r + b ) ^{(n)}}$

With $r = 1, b = 1, n = 2, k = 1$ :

$P = \frac{( 1 2 ) \cdot 1 ^{(1)} \cdot 1 ^{(1)}}{2 ^{(2)}} = \frac{2 \cdot 1 \cdot 1}{2 \cdot 3} = \frac{2}{6} = \frac{1}{3}$

Wrong path walkthrough: A student who treats draws as independent coin flips computes $P (exactly 1 head in 2 flips) = (1 2) (1/2)^{2} = 1/2$ , matching option C. The mistake is at the independence assumption: in a Polya Urn, the second draw's probability depends on the first draw's outcome. The reinforcement mechanism makes draws dependent, reducing the probability of mixed outcomes.

Generalisation: In a Polya Urn, reinforcement amplifies early outcomes, making "all same colour" more likely and "mixed" less likely compared to independent draws.

Verification: The four sequence probabilities sum to $\frac{1}{3} + \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = 1$ . ✓

Answer: B) $\frac{1}{3}$

Question 2 · Engineering Mathematics · 2021_Set2 MCQ

A bag has $r$ red balls and $b$ black balls. All balls are identical except for their colours. In a trial, a ball is randomly drawn from the bag, its colour is noted and the ball is placed back into the bag along with another ball of the same colour. Note that the number of balls in the bag will increase by one, after the trial. A sequence of four such trials is conducted. Which one of the following choices gives the probability of drawing a red ball in the fourth trial?

A.
$\frac{r}{r + b}$
B.
$\frac{r}{r + b + 3}$
C.
$\frac{r + 3}{r + b + 3}$
D.
$(\frac{r}{r + b}) (\frac{r + 1}{r + b + 1}) (\frac{r + 2}{r + b + 2}) (\frac{r + 3}{r + b + 3})$

Correct Answer:

Step-by-Step Solution

Insight: This is a Polya Urn martingale question — the probability of drawing red at any step equals the initial proportion $\frac{r}{r + b}$ .

Exam route: Recognise the Polya Urn mechanism (draw, replace, add one of the same colour). By the martingale property, $P (red at step n) = \frac{r}{r + b}$ for every $n$ . Select option A immediately.

Learning route:

This is a Polya Urn reinforcement process question, identifiable because each draw reinforces the drawn colour by adding one extra ball of that colour.

The key property is the martingale property: let $X_{n} = \frac{r + K _{n}}{r + b + n}$ be the proportion of red balls after $n$ draws, where $K_{n}$ is the number of reds drawn. Then $E [X_{n + 1} ∣ X_{n}] = X_{n}$ . The expected proportion never changes.

Since $E [X_{0}] = \frac{r}{r + b}$ , and the expectation is preserved at every step, the unconditional probability of drawing red at step $n$ is:

$P (red at step n) = E [X_{n - 1}] = \frac{r}{r + b}$

Alternatively, by exchangeability, the joint distribution of draws is symmetric under permutation, so $P (red on 4th draw) = P (red on 1st draw) = \frac{r}{r + b}$ .

Wrong path walkthrough: A student who does not know the martingale property might try option D, computing $\frac{r}{r + b} \cdot \frac{r + 1}{r + b + 1} \cdot \frac{r + 2}{r + b + 2} \cdot \frac{r + 3}{r + b + 3}$ . This is $P (RRRR)$ — the probability of drawing red on all four draws — not the marginal probability at step 4. The mistake breaks at the first multiplication: multiplying conditional probabilities gives the joint, not the marginal.

Generalisation: In any Polya Urn, the marginal probability of drawing a specific colour at any single step equals its initial proportion, regardless of the step number.

Verification: Set $r = 1, b = 1$ . By enumeration over all $2^{4} = 16$ weighted paths, $P (red on 4th) = \frac{1}{2} = \frac{r}{r + b}$ . ✓

Answer: A) $\frac{r}{r + b}$

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

Urn Models and Reinforcement Processes PYQs for GATE CS

Chapter Roadmap: Urn Models and Reinforcement Processes

Your Journey Through Urn Models

Why This Matters for GATE

The Polya Urn: Rich Get Richer

The Polya Urn: Rich Get Richer

Urn Models and Reinforcement Processes: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

More previous year questions (pyqs) in this unit