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JEE MAIN - Mathematics (2025 - 22nd January Morning Shift - No. 11)

A coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma^2$ denote the mean and variance of $X$, then the value of $64\left(\mu+\sigma^2\right)$ is:
64
32
51
48

Explanation

Outcome $x_i$ $p_i$
HHH 0 $\frac{1}{8}$
TTT 0 $\frac{1}{8}$
HHT 1 $\frac{1}{8}$
HTH 1 $\frac{1}{8}$
THH 0 $\frac{1}{8}$
TTH 0 $\frac{1}{8}$
THT 1 $\frac{1}{8}$
HTT 1 $\frac{1}{8}$
$\begin{aligned} & \mu=\sum x_i P_i=\frac{1}{2} \\ & \sigma^2=\sum x_i^2 P_i-\mu^2 \\ & =\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\ & 64\left(\mu+\sigma^2\right)=64\left[\frac{1}{2}+\frac{1}{4}\right] \\ & =64 \times \frac{3}{4}=48\end{aligned}$

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