Question

A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed three times, find the probability distribution of number of tails. Hence, find the mean of the distribution.

Hint:

When one outcome is $k$ times more likely than another, assign probabilities $p$ and $kp$ and solve using $p + kp = 1$.

Answer 1

Let the probability of tail = $p$ and the probability of head = $3p$. Since the total probability must be 1: \[ p + 3p = 1 \Rightarrow 4p = 1 \Rightarrow p = \frac{1}{4} \] Thus, \[ P(\text{tail}) = \frac{1}{4}, \quad P(\text{head}) = \frac{3}{4} \] Let $X$ be the number of tails in 3 tosses. Then $X$ follows a Binomial distribution: \[ X \sim B(n = 3, p = \frac{1}{4}) \] The probability distribution of $X$ is given by: \[ P(X = r) = {3 \choose r} \left(\frac{1}{4}\right)^r \left(\frac{3}{4}\right)^{3 - r}, \quad r = 0, 1, 2, 3 \] Now compute each: - $P(X = 0) = {3 \choose 0} \left(\frac{1}{4}\right)^0 \left(\frac{3}{4}\right)^3 = 1 \cdot 1 \cdot \frac{27}{64} = \frac{27}{64}$ - $P(X = 1) = {3 \choose 1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^2 = 3 \cdot \frac{1}{4} \cdot \frac{9}{16} = \frac{27}{64}$ - $P(X = 2) = {3 \choose 2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^1 = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} = \frac{9}{64}$ - $P(X = 3) = {3 \choose 3} \left(\frac{1}{4}\right)^3 \left(\frac{3}{4}\right)^0 = 1 \cdot \frac{1}{64} \cdot 1 = \frac{1}{64}$ Probability distribution table: \[ \begin{array}{|c|c|} \hline \text{No. of Tails } (X) & P(X) \\ \hline 0 & \frac{27}{64} \\ 1 & \frac{27}{64} \\ 2 & \frac{9}{64} \\ 3 & \frac{1}{64} \\ \hline \end{array} \] Mean of the distribution: \[ \mu = E(X) = \sum X \cdot P(X) = 0 \cdot \frac{27}{64} + 1 \cdot \frac{27}{64} + 2 \cdot \frac{9}{64} + 3 \cdot \frac{1}{64} = \frac{27 + 18 + 3}{64} = \frac{48}{64} = \frac{3}{4} \] % Final Answer Final Answer: The probability distribution is as shown above, and the mean of the distribution is $\frac{3}{4}$.