1. Find the probabilities of Head and Tail

Let $P(H)=p,\, P(T)=q$

Given: head is 3 times as likely as tail, so $p=3q$

Also, $p+q=1$

Thus, $3q+q=1 \Rightarrow 4q=1 \Rightarrow q=\frac14$ and $p=\frac34$

So, $P(H)=\frac34, \qquad P(T)=\frac14$

2. Understand when the experiment stops

The coin is tossed until either:

• a head occurs, or
• three tails occur.

Let $X$ be the number of tosses.

Possible values of $X$ are:

• $1$ : if first toss is $H$
• $2$ : if first toss is $T$ and second toss is $H$
• $3$ : if first two tosses are $TT$ and then third toss is either $H$ or $T$

Why is $X=3$ in both last cases?

• If sequence is $TTH$, then head occurs on 3rd toss.
• If sequence is $TTT$, then three tails occur on 3rd toss.

3. Compute the distribution of $X$

Case 1: $X=1$

This happens if first toss is head. $P(X=1)=P(H)=\frac34$

Case 2: $X=2$

This happens if first toss is tail and second toss is head. $P(X=2)=P(TH)=\frac14\cdot\frac34=\frac{3}{16}$

Case 3: $X=3$

This happens if first two tosses are tails. Then the process must continue to the third toss. $P(X=3)=P(TT)=\left(\frac14\right)^2=\frac{1}{16}$

Check: $\frac34+\frac{3}{16}+\frac{1}{16}=\frac{12}{16}+\frac{3}{16}+\frac{1}{16}=1$

4. Find the mean of $X$

$E(X)=1\cdot P(X=1)+2\cdot P(X=2)+3\cdot P(X=3)$

Substitute values: $E(X)=1\cdot\frac34+2\cdot\frac{3}{16}+3\cdot\frac{1}{16}$

$E(X)=\frac34+\frac{6}{16}+\frac{3}{16}$

Convert $\frac34$ to sixteenths: $\frac34=\frac{12}{16}$

Therefore, $E(X)=\frac{12}{16}+\frac{6}{16}+\frac{3}{16}=\frac{21}{16}$

5. Match with options

$\frac{21}{16}$ corresponds to Option C.