Question

n a poker hand consisting of 5​ cards, find the probability of holding​ (a) 2 jacks​; ​(b) 1 diamond and 4 spades.

Answer 1

Answer:

Given that:

n a poker hand consisting of 5​ cards, find the probability of holding​

(a) 2 jacks​;

The number of ways of being 2 jacks from 4 jacks total is ⁴c₂ = 4!/2! 2!

= 4*3*2/2!*2!

= 4*3/2*1

= 6

The number of ways of selecting 3 other cards from the deck is ⁴⁸c₃ = 48! /3! * 5!

= 48*47*46*45/ 3*2 *45!

= 8* 47*46

= 17296

The number of ways of making this selection follows the multiplication rule

= 6 * 17296

= 103776

The total number of 5-cards poker hand, all of which are equally likely is ⁵²c₅ = 52!/5!47!

= 2598960

so the probability of helding 2 jacks is (⁴c₂ *⁴⁸c₃) / ⁵²c₅

= 103776/2598960

= 0.03992982

(b)  1 diamond and 4 spades.

The number of ways of being 4 spades from 13 spades is ¹³c₄ = 715

The number of ways of selecting 1 diamond from 13 diamond is ¹³c₁ = 13

So, the number of ways of making this selection follows the multiplication rule = 715*13

= 9295

Total number of 5-card power hand is N = ⁵²C₅ = 2598960

Probability 4 spades and 1 diamond is = (¹³c₄*¹³c₁)/⁵²c₅

= 9295/2598960

= 143/ 39984

= 0.00358