Problem Statement
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability distribution of the number of successes $X$. Also find the mean of the distribution.
Verified Solution & Marking Scheme
Probability of single trial success
A doublet occurs when outcomes are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. Total outcomes $= 36$.
$$p = P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}, \quad q = 1 - p = \frac{5}{6}$$
Number of trials $n = 3$. The random variable $X$ can take values $0, 1, 2, 3$.
Compute Probability Distribution P(X = r)
$$P(X = 0) = \binom{3}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^3 = 1 \times 1 \times \frac{125}{216} = \frac{125}{216}$$
$$P(X = 1) = \binom{3}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^2 = 3 \times \frac{1}{6} \times \frac{25}{36} = \frac{75}{216}$$
$$P(X = 2) = \binom{3}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^1 = 3 \times \frac{1}{36} \times \frac{5}{6} = \frac{15}{216}$$
$$P(X = 3) = \binom{3}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^0 = 1 \times \frac{1}{216} \times 1 = \frac{1}{216}$$
Check sum: $\frac{125 + 75 + 15 + 1}{216} = \frac{216}{216} = 1$.
Calculate Mean E(X)
$$\text{Mean } E(X) = \sum x_i P(X = x_i) = 0\left( \frac{125}{216} \right) + 1\left( \frac{75}{216} \right) + 2\left( \frac{15}{216} \right) + 3\left( \frac{1}{216} \right)$$
$$= \frac{0 + 75 + 30 + 3}{216} = \frac{108}{216} = \frac{1}{2} = 0.5$$