Question

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 :

• A. 51
• B. 48
• C. 32
• D. 64

Answer 1

Answer: (B)

48

Answer 2

Solution Explanation:

Let the coin be tossed three times and label the tosses as $T_1, T_2, T_3$. Define indicator variables:

$$I_1 = \begin{cases}1, & \text{if } T_1 = H \text{ and } T_2 = T \\ 0, & \text{otherwise}\end{cases}, \quad I_2 = \begin{cases}1, & \text{if } T_2 = H \text{ and } T_3 = T \\ 0, & \text{otherwise}\end{cases}$$

Then, $X = I_1 + I_2$.

Since the coin is fair:

$$E(I_1) = E(I_2) = P(H \text{ then } T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$$

Thus,

$$\mu = E(X) = E(I_1) + E(I_2) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.$$

For the variance, note:

$$\operatorname{Var}(I_i) = \frac{1}{4}\left(1 - \frac{1}{4}\right) = \frac{3}{16} \quad \text{for } i=1,2.$$

Since $I_1$ and $I_2$ are dependent (they share toss $T_2$):

$$E(I_1I_2) = 0 \quad \text{(because $T_2$ cannot be both $T$ and $H$)},$$ $$\operatorname{Cov}(I_1, I_2) = E(I_1I_2) - E(I_1)E(I_2) = 0 - \frac{1}{4}\cdot\frac{1}{4} = -\frac{1}{16}.$$

Thus,

$$\sigma^2 = \operatorname{Var}(X) = \operatorname{Var}(I_1) + \operatorname{Var}(I_2) + 2\operatorname{Cov}(I_1, I_2) = \frac{3}{16} + \frac{3}{16} - \frac{2}{16} = \frac{4}{16} = \frac{1}{4}.$$

Finally,

$$\mu + \sigma^2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4},$$ $$64(\mu + \sigma^2) = 64 \times \frac{3}{4} = 48.$$

Answer: 48 (Option 2)