Question

Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.

Answer 1

Total number of outcomes when two dice are thrown = 6 x 6, i.e. 36
Let X be the number of doublets in three throws of a pair of dice.
Then, X follows a binomial distribution with n = 3

\[p = P(\text{ Getting a doublet in three throws } ) = \frac{6}{36} = \frac{1}{6} \text{ and } q = \frac{5}{6}\]
\[ \therefore P(X = r) = ^ {3}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{3 - r} , r = 0, 1, 2, 3\]
\[\text{ Mean }\left( \text{ np } \right) = 3\left( \frac{1}{6} \right) = \frac{1}{2}\]

\[\text{ The distribution is as follows } : \]
   X        0             1                 2                 3

\[P(X) \left( \frac{5}{6} \right)^3 \ 3 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{3 - 1} 3 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{3 - 2} \left( \frac{1}{6} \right)^3 \]
\[ \frac{125}{216}            \                     \frac{75}{216} \frac{15}{216} \frac{1}{216}\]