GATE CS 2026 Set 1 Q1 — Probability: Random Variables and Distributions

We are given an urn initially containing $1$ red ball ($\text{R}$) and $1$ blue ball ($\text{B}$). At each step, we draw a ball, and return it to the urn along with an additional ball of the same color.

Initially, the total number of balls is $2$. After two steps, $2$ more balls will be added, making the total number of balls equal to $4$. For there to be an equal number of red and blue balls after two steps, the urn must contain exactly $2\text{ R}$ and $2\text{ B}$ balls.

Let us analyze the two possible sequences of draws that result in $2\text{ R}$ and $2\text{ B}$ balls:

1. Sequence 1: Red then Blue ($\text{R} \to \text{B}$)
   • Step 1: Draw a Red ball. The probability is $\frac{1}{2}$. The urn now contains $2\text{ R}$ and $1\text{ B}$ (total $3$ balls).
   • Step 2: Draw a Blue ball. The probability is $\frac{1}{3}$.
   • The probability of this sequence is: $$P(\text{R} \to \text{B}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$$
2. Sequence 2: Blue then Red ($\text{B} \to \text{R}$)
   • Step 1: Draw a Blue ball. The probability is $\frac{1}{2}$. The urn now contains $1\text{ R}$ and $2\text{ B}$ (total $3$ balls).
   • Step 2: Draw a Red ball. The probability is $\frac{1}{3}$.
   • The probability of this sequence is: $$P(\text{B} \to \text{R}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$$

Since these two sequences are mutually exclusive, the total probability of having an equal number of red and blue balls after two steps is:

Thus, the correct option is B.