Example 15
Find the probability that when a hand of 7 cards is drawn from a well
shuffled deck of 52 cards, it contains
(i) all Kings
7 cards are to be chosen from 52 cards
Total number of combinations (hands) possible = 52C7
= 52!/7!(52 −7)! = 52!/7!45!
Let A be the event that all kings are selected
There are only 4 kings in a pack of 52 cards
Hence if 7 cards are chosen
4 kings to be chosen out of 4 and 3 others to be chosen out of remaining 48
Hence total number of combinations
n(A) = 4C4 × 48C3
= 4!/4!0! × 48!/3!(48 −3)! = 1× 48/(3! 45!) = 48!/(3! 45!)
Hence
P (A) = (𝑛(𝐴))/(𝑛(𝑆))
= 48!/(3! 45!) ÷ 52!/(7! 45!)
= (48! × 7!)/(3! × 52!) = 𝟏/𝟕𝟕𝟑𝟓
Example 15
Find the probability that when a hand of 7 cards is drawn from a well
shuffled deck of 52 cards, it contains
(ii) 3 Kings
Let B be that event that 3 king are selected
There are only 4 king in a pack of 52 cards
Hence if 7 cards are chosen,
3 king to be chosen out of 4 and 4 other to be chosen out of remaining 48
Hence, number of combination
n(B) = 4C3 × 48C4
= 4!/3!(4 −3)! × 48!/4!(48 −4)!
= 4 × 48!/4!44!
Hence,
P(B) = (𝑛(𝐵))/(𝑛(𝑆))
= (4 × 48!/4!44!)/(52!/7!45!)
= (4 × 48!)/4!44! × 7!45!/52!
= (4 × 48! × 7! × 45!)/(4! × 44! × 52!)
= (4 × 48! × 7! × 45!)/(4! × 44! × 52 × 51 × 50 × 49 × 48!)
= (4 × 7! × 45)/(4! × 52 × 51 × 50 × 49 )
= (4 × 7 × 6 × 5 × 4! × 45)/( 52 × 51 × 50 × 49 × 4!)
= 𝟗/𝟏𝟓𝟒𝟕
Example 15
Find the probability that when a hand of 7 cards is drawn from a well
shuffled deck of 52 cards, it contains
(iii) at least 3 Kings.
Atleast 3 kings are selected means
either 3 kings are selected or 4 kings are selected
So, P(at least 3 king) = P(3 King) + ( 4 King)
We know that,
P(3 Kings) = 9/1547
P(4 Kings) = 1/7735
(calculated in (ii) part)
(calculated in (i) part)
P(at least 3 king) = 9/1547 + 1/7735
= 𝟒𝟔/𝟕𝟕𝟑𝟓

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.