Question 2.33: What is the probability that a poker hand contains 4 picture......

A poker hand can be selected in {\binom{52}{5}} ways. The event described, call it A, consists of the following number of sample points: n(A) = n(J_{2})+n(J_{3})+n(J_{4}), where ~{J}_{i}=~ “the poker hand contains exactly i Jacks,” i=2,3,4.~ But

n(J_{2})={\binom{4}{2}}{\binom{8}{2}}{\binom{40}{1}},\quad n(J_{3})={\binom{4}{3}}{\binom{8}{1}}{\binom{40}{1}},\quad n(J_{3})={\binom{4}{4}}{\binom{8}{0}}{\binom{40}{1}},

so that

P(A)={\frac{\left[\binom {4}{2}\binom{8}{2}+\binom{4}{3}\binom{8}{1}+\binom{4}{4}\binom{8}{0}\right]\binom{40}{1}}{\binom{52}{5}}}={\frac{8,040}{2,598,960}}\simeq0.003.

(For the calculation of \textstyle{\binom{52}{5}} see Example 29(ix).)