QJEEM23Ma4d92: JEE Main 2023 Mathematics - Class 12 JEEM23Ma4d92 Solution: A coin is biased so that the head is 3 times as likely to occur …

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:

$\begin{aligned}E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\& = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\& = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\& = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\& = 3 \cdot \left(\frac{7}{16}\right) \\\\& = \frac{21}{16}.\end{aligned}$