Answer to Question #139079 in Statistics and Probability for qwerty

Question

2. In a poker hand consisting of 5 cards, find the probability of holding
a. 3 kings
b. 4 hearts and 1 diamonds

Answer 1

A poker hand consists of 5 cards. The sample space has $\dbinom{52}{5}$ possible outcomes.

a. Choose 3 kings from 4 kings. From the remaining $52-4=48$ cards choose 2 non-kings to make a five-card poker hand

$$P(3\ kings)=\dfrac{\dbinom{4}{3}\dbinom{52-4}{5-3}}{\dbinom{52}{5}}=$$

$$=\dfrac{\dfrac{4!}{3!(4-3)!}\cdot\dfrac{48!}{2!(48-2)!}}{\dfrac{52!}{5!(52-5)!}}=$$

$$=\dfrac{2(48)(47)(120)}{52(51)(50)(49)(48)}=\dfrac{94}{54145}\approx0.001736$$

b. There are total 13 hearts and 13 diamonds.

Choose 4 hearts from 13 hearts

$$\dbinom{13}{4}$$

Choose 1 diamonds from 13 diamonds

$$\dbinom{13}{1}$$

$$P(4\ hearts\ and\ 1\ diamonds)=\dfrac{\dbinom{13}{4}\dbinom{13}{1}}{\dbinom{52}{5}}=$$

$$=\dfrac{\dfrac{13!}{4!(13-4)!}\cdot\dfrac{13!}{1!(13-1)!}}{\dfrac{52!}{5!(52-5)!}}=$$

$$=\dfrac{13(12)(11)(10)(5)(13)}{52(51)(50)(49)(48)}=\dfrac{143}{39984}\approx0.003576$$