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:

Draw a probability tree for the two steps: after step 1 the urn is either (2 red, 1 blue) or (1 red, 2 blue), and in each case exactly one of the two step-2 outcomes restores equal counts.

Answer 1

The Correct Option is B

Step 1: Understand the process (this is a Polya urn model).
Start with 1 red (R) and 1 blue (B) ball, total 2 balls. At each step, one ball is drawn uniformly at random, then it is returned to the urn together with one extra ball of the same color, so the urn grows by exactly one ball per step. After step 1 the urn has 3 balls, and after step 2 the urn has 4 balls.

Step 2: List the two possible outcomes of step 1.
With probability \(\frac{1}{2}\), a red ball is drawn, so the urn becomes 2 red, 1 blue (3 balls total).
With probability \(\frac{1}{2}\), a blue ball is drawn, so the urn becomes 1 red, 2 blue (3 balls total).

Step 3: For each case, work out what step 2 must draw to make the counts equal (2 red, 2 blue after step 2, since 4 balls total split evenly is 2 and 2).
Case A, urn is (2 red, 1 blue): drawing red gives (3 red, 1 blue), unequal; drawing blue gives (2 red, 2 blue), equal. The probability of drawing blue here is \(\frac{1}{3}\) (1 blue ball out of 3).
Case B, urn is (1 red, 2 blue): drawing blue gives (1 red, 3 blue), unequal; drawing red gives (2 red, 2 blue), equal. The probability of drawing red here is \(\frac{1}{3}\) (1 red ball out of 3).

Step 4: Combine the two cases using the law of total probability.
\[ P(\text{equal after 2 steps}) = P(\text{Case A}) \times P(\text{blue in step 2} \mid \text{Case A}) + P(\text{Case B}) \times P(\text{red in step 2} \mid \text{Case B}) \]
\[ = \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

Step 5: Check against the options.
\(\frac{1}{3}\) matches option (B). Option (A) \(\frac{1}{4}\) does not arise from any valid path probability in this two step tree. Option (C) \(\frac{1}{2}\) would only be reached if every step-2 draw led to equal counts, which is false since each case has exactly one of its two branches (probability \(\frac{2}{3}\) combined) leading to unequal counts. Option (D) \(\frac{2}{3}\) is actually the probability of the complementary event, ending with unequal counts after two steps.
\[ \boxed{\dfrac{1}{3}} \]