A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $\mathrm{X}$ denotes the number of tosses of the coin, then the mean of $\mathrm{X}$ is :

The given probabilities for getting a head (H) and a tail (T) are as follows: <br/><br/>$P(H) = \frac{3}{4}, \quad P(T) = \frac{1}{4}$ <br/><br/>The random variable X can take the values 1, 2, or 3. These correspond to the following events: <br/><br/>- X = 1 : A head is obtained on the first toss. This happens with probability $P(H) = \frac{3}{4}$. <br/><br/>- X = 2 : A tail is obtained on the first toss and a head on the second. This happens with probability $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$. <br/><br/>- X = 3 : Either two tails and then a head are obtained, or three tails are obtained. <br/> This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$. <br/><br/>Now, we calculate the mean (expected value) of X: <br/><br/>$$ \begin{aligned} E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\ & = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\ & = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\ & = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\ & = 3 \cdot \left(\frac{7}{16}\right) \\\\ & = \frac{21}{16}. \end{aligned} $$