Question

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.

Answer 1

Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then,

\[p = \frac{6}{36} = \frac{1}{6}\]
\[q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}\]

p = Let X be the number of getting doublets in 4 throws of a pair of dice. Then, X follows a binomial distribution with n =4,

\[p = \frac{1}{6}\text{ and } q = \frac{5}{6}\]
\[P(X = r) = \text{Probability of getting r doublets} \]
\[P(X = r) = ^{4}{}{C}_r (\frac{1}{6} )^r (\frac{5}{6} )^{4 - r} ; r = 0, 1, 2, 3, 4\]
\[\text{ If } X = 0, \text{ then } P(X = 0) = ^{4}{}{C}_0 (\frac{1}{6} )^0 (\frac{5}{6} )^{4 - r} \]
\[ \Rightarrow P = \left( \frac{5}{6} \right)^4 \]
\[\text{ If } X = 1, \text{ then } P = ^{4}{}{C}_1 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{4 - 1} = \frac{2}{3} \left( \frac{5}{6} \right)^3 \]
\[\text{ If } X = 2, \text{ then}  P = ^{4}{}{C}_2 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{4 - 2} = \frac{1}{6} \left( \frac{5}{6} \right)^2 \]
\[\text{ If } X = 3, \text{ then } P = ^{4}{}{C}_3 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{4 - 3} = \frac{10}{3} \left( \frac{1}{6} \right)^3 \]

\[\text{ If } X = 4, \text{ then } P = ^{4}{}{C}_4 (\frac{1}{6} )^4 (\frac{5}{6} )^{4 - 4} = \left( \frac{1}{6} \right)^4 = \frac{1}{1296}\]

    X      0        1          2      3       4
\[P(X) (\frac{5}{6} )^4 \frac{2}{3}(\frac{5}{6} )^3 \frac{25}{216} \frac{5}{324} \frac{1}{1296}\]