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 $64(\mu+\sigma^{2})$ is:
Step-by-Step Solution
Key Concept: Enumerate the $8$ equally likely outcomes and count HT-adjacencies in each. The variable $X$ turns out to be Bernoulli (only values $0$ or $1$ — at most one HT in $3$ tosses).
Outcomes: HHH$\to 0$, HHT$\to 1$, HTH$\to 1$, HTT$\to 1$, THH$\to 0$, THT$\to 1$, TTH$\to 0$, TTT$\to 0.$
So $X=1$ in $4$ cases and $X=0$ in $4$ cases: $P(X=1)=\tfrac{1}{2},\,P(X=0)=\tfrac{1}{2}.$
$\mu=\tfrac{1}{2},\ E[X^{2}]=\tfrac{1}{2},\ \sigma^{2}=\tfrac{1}{2}-\tfrac{1}{4}=\tfrac{1}{4}.$
$64(\mu+\sigma^{2})=64\cdot\tfrac{3}{4}=48.$
Correct Answer: 4