A coin is tossed three times. Let X denote the number of times a tail follows a head. If mu and sigma^2 denotes the mean and variance of X, then the value of 64( mu + sigma^2) is : (1) 51 (2) 48 (3) 32 (4) 64

Key ideas you must know

1. Indicator variable: Define a small 0–1 variable that is 1 if the event happens, 0 if it does not. They make means and variances easy because they turn complicated counts into simple sums.
2. Expected value of an indicator: If $I$ is 1 when something happens with probability $p$ and 0 otherwise, then $\mathbb{E}[I]=p$.
3. Variance of a sum:
   $\operatorname{Var}(I_1+I_2)=\operatorname{Var}(I_1)+\operatorname{Var}(I_2)+2\,\operatorname{Cov}(I_1,I_2)$
   Here covariance is needed because the two pairs share the middle toss, so the two indicators are not independent.

Logical chain a student should follow

1. Break the count into pairs: For three tosses there are only two adjacent pairs: (1,2) and (2,3).
2. Define $I_1$ = 1 if pair (1,2) is HT, else 0. $I_2$ = 1 if pair (2,3) is HT, else 0. Then $X=I_1+I_2$
3. Find each single probability: $P(HT)=\tfrac12\times\tfrac12=\tfrac14$.
4. Compute variances of indicators: For an indicator with $p=\tfrac14$, $\operatorname{Var}(I)=p(1-p)=\tfrac14\times\tfrac34=\tfrac{3}{16}.$
5. Check if both events can occur together: For both indicators to be 1, tosses must be HT and HT simultaneously, which is impossible because the middle toss can’t be both H and T. Hence $I_1I_2=0$ always.
6. Use that to find covariance: $\operatorname{Cov}(I_1,I_2) = \mathbb{E}[I_1I_2]-\mathbb{E}[I_1]\mathbb{E}[I_2] = 0 - \left(\tfrac14\right)^2 = -\tfrac{1}{16}$.
7. Assemble mean and variance and finish the arithmetic.

This systematic approach makes the calculation tidy and avoids mistake‐prone enumeration of all eight possible H/T sequences.

What the question says

We toss a coin three times and look at the order of Heads (H) and Tails (T).

• We only care about places where a Tail comes right after a Head.
• We count how many times that happens and call that number X.

Then we have to find the average value (mean $\mu$) and the spread (variance $\sigma^2$) of that count, add them, multiply by 64, and pick the right option.

Picture it like story‐telling

Imagine you say three words in a row. Every time you say “Hi” (H) and then immediately say “Ta‐da” (T), you shout "Got one!". At the end you count how many "Got ones" you had. You want to know, on average, how many times that shout happens if you repeat the game many, many times.

Step 1 – Define indicators

Let $I_1 = \begin{cases}1, & \text{if toss 1 = H and toss 2 = T}\\0, & \text{otherwise}\end{cases}$ $I_2 = \begin{cases}1, & \text{if toss 2 = H and toss 3 = T}\\0, & \text{otherwise}\end{cases}$ Then $X = I_1 + I_2$

Step 2 – Mean of each indicator

Probability that a pair is HT: $P(HT) = \frac12 \times \frac12 = \frac14$ Hence $\mathbb{E}[I_1] = \mathbb{E}[I_2] = \frac14$ So the mean of $X$ is $\mu = \mathbb{E}[X] = \mathbb{E}[I_1] + \mathbb{E}[I_2] = 2 \times \frac14 = \frac12$

Step 3 – Variance of indicators

For an indicator, $\operatorname{Var}(I_k) = p(1-p) = \frac14\left(1-\frac14\right) = \frac{3}{16}$ Thus $\operatorname{Var}(I_1) + \operatorname{Var}(I_2) = 2 \times \frac{3}{16} = \frac38$

Step 4 – Covariance term

Both $I_1$ and $I_2$ can be 1 only if the pattern is HT and HT overlapping, which is impossible. Therefore $I_1 I_2 = 0 \;\text{always} \;\Rightarrow\; \mathbb{E}[I_1 I_2] = 0$ So $\operatorname{Cov}(I_1,I_2) = 0 - \left(\frac14\right)^2 = -\frac{1}{16}$

Step 5 – Variance of $X$

$\sigma^2 = \operatorname{Var}(X) = \frac38 + 2\left(-\frac{1}{16}\right) = \frac38 - \frac18 = \frac14$

Step 6 – Final expression

$\mu + \sigma^2 = \frac12 + \frac14 = \frac34$ Multiply by 64: $64\left(\mu + \sigma^2\right) = 64 \times \frac34 = 48$

Hence the correct option is (2) 48.