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:
- (1) 51
- (2) 48
- (3) 32
- (4) 64
Solution:
Sample space for tossing a coin 3 times ($2^3 = 8$ outcomes):
$$ S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} $$
Random variable $X$ represents the occurrence of “Tail immediately following a Head” (HT sequence).
- X = 0: HHH, THH, TTH, TTT (4 outcomes)
- X = 1: HHT, HTH, HTT, THT (4 outcomes)
Probability distribution:
$$ P(X=0) = \frac{4}{8} = \frac{1}{2} $$
$$ P(X=1) = \frac{4}{8} = \frac{1}{2} $$
Calculation of Mean ($\mu$) and Variance ($\sigma^2$):
$$ \mu = \sum x P(x) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2} $$
$$ E(X^2) = \sum x^2 P(x) = 0^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{2} = \frac{1}{2} $$
$$ \sigma^2 = E(X^2) – (\mu)^2 = \frac{1}{2} – \left(\frac{1}{2}\right)^2 = \frac{1}{2} – \frac{1}{4} = \frac{1}{4} $$
Required value:
$$ 64(\mu + \sigma^2) = 64\left(\frac{1}{2} + \frac{1}{4}\right) = 64\left(\frac{3}{4}\right) = 48 $$
Ans. (2)