Answer to Question #139079 in Statistics and Probability for qwerty

Question #139079
2. In a poker hand consisting of 5 cards, find the probability of holding
a. 3 kings
b. 4 hearts and 1 diamonds
1
Expert's answer
2020-10-19T17:46:54-0400

A poker hand consists of 5 cards. The sample space has (525)\dbinom{52}{5} possible outcomes.

a. Choose 3 kings from 4 kings. From the remaining 524=4852-4=48 cards choose 2 non-kings to make a five-card poker hand

P(3 kings)=(43)(52453)(525)=P(3\ kings)=\dfrac{\dbinom{4}{3}\dbinom{52-4}{5-3}}{\dbinom{52}{5}}=

=4!3!(43)!48!2!(482)!52!5!(525)!==\dfrac{\dfrac{4!}{3!(4-3)!}\cdot\dfrac{48!}{2!(48-2)!}}{\dfrac{52!}{5!(52-5)!}}=

=2(48)(47)(120)52(51)(50)(49)(48)=94541450.001736=\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


(134)\dbinom{13}{4}

Choose 1 diamonds from 13 diamonds


(131)\dbinom{13}{1}


P(4 hearts and 1 diamonds)=(134)(131)(525)=P(4\ hearts\ and\ 1\ diamonds)=\dfrac{\dbinom{13}{4}\dbinom{13}{1}}{\dbinom{52}{5}}=

=13!4!(134)!13!1!(131)!52!5!(525)!==\dfrac{\dfrac{13!}{4!(13-4)!}\cdot\dfrac{13!}{1!(13-1)!}}{\dfrac{52!}{5!(52-5)!}}=

=13(12)(11)(10)(5)(13)52(51)(50)(49)(48)=143399840.003576=\dfrac{13(12)(11)(10)(5)(13)}{52(51)(50)(49)(48)}=\dfrac{143}{39984}\approx0.003576


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment