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 :

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

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 :

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