Question

Find the probability that when a hand of 7 cards are drawn from the well-shuffled deck of 52 cards, it contains

(i) all kings (ii) 3 kings​

Answer 1

$\huge\underline{\mathbb\red{❥︎A}\green{N}\mathbb\blue{S}\purple{W}\mathbb\orange{E}\pink{R}}\:$

Solution:

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(i) To find the probability that all the cards are kings:

If 7 cards are chosen from the pack of 52 cards

Then the total number of combinations possible are: 52C7

= 52!/[7! (52-7)!]

= 52!/ (7! 45!)

Assume that A be the event that all the kings are selected

We know that there are only 4 kings in the pack of 52 cards

Thus, if 7 cards are chosen, 4 kings are chosen out of 4, and 3 should be chosen form the 48 remaining cards.

Therefore, the total number of combinations is:

n(A) = 4C4 x48C3

= [4!/4!0! ] x [48!/3!(48-3)!]

= 1 x [48!/3! 45!]

= 48!/3! 45!

Therefore, P(A) = n(A)/n(S)

= [48!/3! 45!] ÷[52!/ (7! 45!]

= [48! x 7!] ÷ [3!x 52!]

= 1/7735

Therefore, the probability of getting all the 7 cards are kings is 1/7735

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(ii) To find the probability that 3 cards are kings:

Assume that B be the event that 3 kings are selected.

Thus, if 7 cards are chosen, 3 kings are chosen out of 4, and 4 cards should be chosen form the 48 remaining cards.

Therefore, the total number of combinations is:

n(B) = 4C3 x48C4

= [4!/3!(4-3)! ] x [48!/4!(48-4)!]

= 4 x [48!/4! 44!]

= 48!/3! 45!

Therefore, P(B) = n(B)/n(S)

= [4 x48!/4! 44!] ÷[52!/ (7! 45!]

= 9/1547

Therefore, the probability of getting 3 kings is 9/1547

$\huge\sf\underline{\underline{\red{❥︎ Thanks}}}$

Answer 2

Answer:

Hi!!!

refer attachments!!!!

hope it helps!!!!