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Question
Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribtution.
Solution
\[\text{ We have } , \]
\[p = \text{ probability of getting a spade in a draw } = \frac{13}{52} = \frac{1}{4} \text{ and } \]
\[q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}\]
\[\text{ Let X denote a success of getting a spade in a throw . Then,} \]
\[\text{ X follows binomial distribution with parameters n = 3 and } p = \frac{1}{4}\]
\[ \therefore P\left( X = r \right) = ^{3}{}{C}_r p^r q^\left( 3 - r \right) =^{3}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^\left( 3 - r \right) = \frac{^{3}{}{C}_r 3^\left( 3 - r \right)}{4^3} = \frac{27}{64}\left( \frac{^{3}{}{C}_r}{3^r} \right), \text{ where } , r = 0, 1, 2, 3\]
\[\text{ So, the probability distribution of X is given by: } \]
\[P\left( X = r \right) = \frac{27}{64}\left( \frac{^{3}{}{C}_r}{3^r} \right), \text{ where} , r = 0, 1, 2, 3\]
\[\text{ Now } , \]
\[\text{ Mean } , E\left( X \right) = np = 3 \times \frac{1}{4} = \frac{3}{4}\]
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