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

							Dear Student

(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) =4C4x48C3

= [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

(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) =4C3x48C4

= [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

Thanks