Tom K. answered 12/16/20
Knowledgeable and Friendly Math and Statistics Tutor
a) P(at least one is a jack or queen) = 1 - P(all are not jacks or queens)
As we are drawing 5 cards from 52, and 44 are not jack or queen), this equals
1 - C(44,5)/C(52,5) = 27017/46410 = .582137
b) P(first picked is a club) = 13/52 = 1/4 = .25
c) 16 of 52 cards are face cards, and 36 not face cards)
P(at least 1 face card) = 1 - P(no face cards) = 1 - C(36,5)/C(52,5) = 389/455 = .854945
d)P(first 2 cards are spades) = P(13,2)/P(52,2) = 13*12/(52*51) = 1/17 = .058824
e)P(both a jack and queen) = 1 - P(jack or not jack or queen) - P(queen or not jack or queen) + P(not jack and not queen) = 1 - (2 C(48,5) + 2 C(44,5))/C(52,5) = 6509/64974 = .100179
Tom K.
It would be 1 - C(39,5)/C(52,5) = 7411/9520 = .7785.12/16/20
Swag M.
For b I'm trying to find the probability that at least 1 club card is within the 5 randomly drawn cards12/16/20