Question

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. 1/4
• B. 1/3
• C. 1/2
• D. 2/3

Hint:

Track the urn state after each draw: starting at (1R,1B), find the probability of each possible state after step 1, then compute the probability of drawing the opposite colour in step 2 (which restores balance) for each branch, and sum.

Answer 1

The Correct Option is B

This is a classic Pólya urn problem where we track how the composition of the urn changes over two random draws.

Step 1: Initial setup.

The urn starts with 1 red (R) and 1 blue (B) ball, so the total is 2 balls. At each step, a ball is drawn uniformly at random and replaced along with another ball of the same colour, so the total number of balls increases by 1 after each step.

Step 2: Enumerate outcomes of the first draw.

With probability \( \frac{1}{2} \), a red ball is drawn. It is returned along with another red ball, giving the state \( (2R, 1B) \), total 3 balls.

With probability \( \frac{1}{2} \), a blue ball is drawn. It is returned along with another blue ball, giving the state \( (1R, 2B) \), total 3 balls.

Step 3: Analyze the second draw from state (2R, 1B).

Here \( P(\text{red}) = \frac{2}{3} \) and \( P(\text{blue}) = \frac{1}{3} \). Drawing another red gives \( (3R,1B) \) - unequal. Drawing a blue gives \( (2R,2B) \) - equal! So the probability of reaching equal counts from this branch is \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).

Step 4: Analyze the second draw from state (1R, 2B).

Here \( P(\text{blue}) = \frac{2}{3} \) and \( P(\text{red}) = \frac{1}{3} \). Drawing another blue gives \( (1R,3B) \) - unequal. Drawing a red gives \( (2R,2B) \) - equal! So the probability of reaching equal counts from this branch is \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).

Step 5: Add the two mutually exclusive branches.

\[ P(\text{equal after 2 steps}) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

Final Answer: The probability is \( \frac{1}{3} \), option (B).