A coin is tossed three times. Let X denote the number of times a tail follows a head. If and ^2 denote the mean and vari

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A coin is tossed three times. Let X denote the number of times a tail follows a head. If and ^2 denote the mean and variance of X, then the value of 64( + ^2) is:

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Given: A coin is tossed three times. Let $X$ be the number of times a tail follows a head.

Find: The value of $64(\mu + \sigma^2)$ , where $\mu = E(X)$ and $\sigma^2 = \operatorname{Var}(X)$ .

List all possible outcomes and the corresponding value of $X$ :

$HHH \to 0$
$HHT \to 1$
$HTH \to 1$
$HTT \to 1$
$THH \to 0$
$THT \to 1$
$TTH \to 0$
$TTT \to 0$

Thus, the probability distribution is:

P(X=0) = \frac{4}{8} = \frac{1}{2}, \qquad P(X=1) = \frac{4}{8} = \frac{1}{2}

A two-row probability distribution table showing x_i values 0 and 1, each with probability one-half.

Now compute the mean:

\mu = \sum x_i P_i = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2}

Compute $E(X^2)$ :

E(X^2) = 0^2 \cdot \frac{1}{2} + 1^2 \cdot \frac{1}{2} = \frac{1}{2}

Hence, the variance is:

\sigma^2 = E(X^2) - \mu^2 = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}

Finally,

64(\mu + \sigma^2) = 64\left(\frac{1}{2} + \frac{1}{4}\right) = 64\left(\frac{3}{4}\right) = 48

Therefore, the correct option is B.

Given: There are 8 equally likely outcomes when a coin is tossed three times. Find: Mean and variance of X, then evaluate 64( + ^2). From the outcomes, X=0 occurs in 4 cases and X=1 occurs in 4 cases. So: P(X=0) = 12, P(X=1) = 12 Mean: E(X) = 0 12 + 1 12 = 12 Variance: Var(X) = E(X^2) - [E(X)]^2 with E(X^2) = 0^2 12 + 1^2 12 = 12 Therefore, ^2 = 12 - 14 = 14 Now substitute: 64( + ^2) = 64 ( 12 + 14) = 48 Therefore, the required value is 48, so the correct option is B.

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