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

Think of this problem as tracking a sequence of two colour draws under Pólya's urn replacement rule, where after drawing a ball, one extra ball of the same colour is added back.

Step 1: List possible two-draw sequences. The colour drawn on step 1 can be R or B, and the colour drawn on step 2 can also be R or B, giving four sequences: RR, RB, BR, BB.

Step 2: Compute the probability of each sequence.

Sequence RR: \(P = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\) (after first R, urn is 2R,1B, so second R has probability 2/3). This leaves the urn unbalanced (3R,1B).

Sequence RB: \(P = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). This leaves the urn balanced (2R,2B).

Sequence BR: \(P = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). This leaves the urn balanced (2R,2B).

Sequence BB: \(P = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\). This leaves the urn unbalanced (1R,3B).

Step 3: Identify the "balanced" outcomes. Only the sequences where the two draws have different colours (RB or BR) leave 2 red and 2 blue balls in the urn - a "colour switch" between step 1 and step 2.

Step 4: Add the favourable probabilities.

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

Note the elegant symmetry: swapping "red" and "blue" throughout leaves the urn dynamics unchanged, which is exactly why the RB and BR branches contribute equal probability.

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