GATE CS 2026 Set 2 Q53 — Conditional Probability and Bayes Theorem

We want to find the conditional probability $P(E_1 \mid E_2)$. Let the outcomes of the six tosses be $T_1, T_2, T_3, T_4, T_5, T_6 \in \{H, T\}$.

The events are defined as:

• $E_1$: At least two heads among $\{T_2, T_4, T_6\}$.
• $E_2$: Exactly two heads and two tails among $\{T_1, T_2, T_3, T_5\}$.

Since $T_2$ is the only common coin toss between $E_1$ and $E_2$, we can partition the conditional probability based on the value of $T_2$:

Step 1: Find $P(T_2 = H \mid E_2)$ and $P(T_2 = T \mid E_2)$

For $E_2$ to occur, we need exactly $2$ heads out of $\{T_1, T_2, T_3, T_5\}$:

• If $T_2 = H$, we need exactly $1$ head from $\{T_1, T_3, T_5\}$, which can happen in $\binom{3}{1} = 3$ ways.
• If $T_2 = T$, we need exactly $2$ heads from $\{T_1, T_3, T_5\}$, which can happen in $\binom{3}{2} = 3$ ways.

Since both cases are equally likely, $P(T_2 = H \mid E_2) = P(T_2 = T \mid E_2) = 0.5$.

Step 2: Find the conditional probabilities of $E_1$

Given $T_2$, the event $E_1$ depends only on the independent tosses $T_4$ and $T_6$:

• If $T_2 = H$, then $E_1$ (at least two heads in $\{T_2, T_4, T_6\}$) requires at least one head in $\{T_4, T_6\}$: $$P(E_1 \mid T_2 = H, E_2) = 1 - P(T_4 = T, T_6 = T) = 1 - \frac{1}{4} = \frac{3}{4}$$
• If $T_2 = T$, then $E_1$ requires both $T_4$ and $T_6$ to be heads: $$P(E_1 \mid T_2 = T, E_2) = P(T_4 = H, T_6 = H) = \frac{1}{4}$$

Step 3: Calculate the final probability