Question

Find the probability distribution of the number of doublets in 4 throws of a pair of dice.

 

Answer 1

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

\[p = \text{ No of getting } (1, 1)(2, 2) . . . (6, 6) = \frac{6}{36} = \frac{1}{6}\]
\[q = 1 - p = \frac{5}{6}\]
\[P(X = r) = ^ {4}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{4 - r} , r = 0, 1, 2, 3, 4\]
\[P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{4 - 0} \]
\[P(X = 1) =^{4}{}{C}_1 \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^{4 - 1} \]
\[P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^{4 - 2} \]
\[P(X = 3) = ^{4}{}{C}_3 \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^{4 - 3} \]
\[P(X = 4) = ^{4}{}{C}_4 \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^{4 - 4} \]
\[\text{ The distribution is as follows } . \]
   X        0         1        2        3        4 
\[P(X)       \          \frac{625}{1296} \frac{500}{1296} \frac{150}{1296} \frac{20}{1296} \frac{1}{1296}\]