Question

52 well-shuffled playing cards are distributed at random among 4 player's.Find the probability that a hand contain

(i) 3 aces

(ii) 2 kings and one queen

(iii) No cards of diamond

(iv) 5 pictured cards .

Answer 1

$$52/4=13$$

There are $\dbinom{52}{13}=\dfrac{52!}{13!(52-13)!}=635,013,559,600$ ways to distribute 13 cards to each of four people.

(i)

$$P(3\ aces)=\dfrac{\dbinom{4}{3}\dbinom{52-4}{13-3}}{\dbinom{52}{13}}$$

$$=\dfrac{4(6,540,715,896)}{635,013,559,600}=0.0412$$

(ii)

$$P(2\ kings\ and\ 1\ queen)=\dfrac{\dbinom{4}{2}\dbinom{4}{1}\dbinom{52-8}{13-3}}{\dbinom{52}{13}}$$

$$=\dfrac{6(4)(2,481,256,778)}{635,013,559,600}=0.0938$$

(iii)

$$P(No\ diamonds)=\dfrac{\dbinom{13}{0}\dbinom{52-13}{13-0}}{\dbinom{52}{13}}$$

$$=\dfrac{1(8,122,425,444)}{635,013,559,600}=0.0128$$

(iv)

$$P(5\ picture\ cards)=\dfrac{\dbinom{12}{5}\dbinom{52-12}{13-5}}{\dbinom{52}{13}}$$

$$=\dfrac{792(76,904,685)}{635,013,559,600}=0.0959$$