JEE MAIN - Mathematics (2023 - 13th April Morning Shift - No. 6)
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 :
$$\frac{81}{64}$$
$$\frac{37}{16}$$
$$\frac{21}{16}$$
$$\frac{15}{16}$$
Explanation
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 $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$.
- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.
This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$.
Now, we calculate the mean (expected value) of X:
$$ \begin{aligned} E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\ & = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\ & = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\ & = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\ & = 3 \cdot \left(\frac{7}{16}\right) \\\\ & = \frac{21}{16}. \end{aligned} $$
$$ P(H) = \frac{3}{4}, \quad P(T) = \frac{1}{4} $$
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 $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$.
- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.
This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$.
Now, we calculate the mean (expected value) of X:
$$ \begin{aligned} E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\ & = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\ & = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\ & = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\ & = 3 \cdot \left(\frac{7}{16}\right) \\\\ & = \frac{21}{16}. \end{aligned} $$
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