Shuffle a standard deck of 52 cards throughly (so that all 52! possible ordering

Question

Shuffle a standard deck of 52 cards throughly (so that all 52! possible orderings are equally likely). Then deal the top 4 cards, in order. Find the probability that

(a). all 4 cards are red

(b) the first 2 cards are red and the last 2 cards are black

(c) there are exactly 2 red cards and 2 black cards among the 4 cards

(d) the first card is black and the second card is a spade

(e) the second card is a spade, given that the first card is black

(f) you get one card of each suit (so one spade, one club, one heart, and one diamond)

Explanation / Answer

(a) Here cards are shuffled properly

Pr(all 4 cards are red) = (26/52) * (25/51) * (24/50) * (23/49) = 26P4/52P4 = 0.0552

(b) Pr(first 2 cards are red and last 2 cards are black) = (26/52) * (25/51) * (26/50) * (25/49) = 0.065

(c) Pr(Exxctly 2 red cards and 2 black cards) = 4C2 (26/52) * (25/51) * (26/52) * (25/51) = 0.3902

(d) Pr(First card is black and second card is spade) = Pr(First card is spade and second card is spade) + Pr(First card is club and second card is spade) = 13/52 * 12/51 + 13/52 * 13/51 = 0.1225

(e) Pr(Second card is spade l first card is black)

so as first card is black so there are 51 cards remaining out of which 26 are red and 25 are blakc

Pr(First card is black) = Pr(First card is club) + Pr(first card is spade)

Pr(Second card is spade  l first card is black) = (13/51) * (1/4) + (12/51) * (1/4) = 0.1225

(f) Pr(One card of each suit) = 4! * (13/52) * (13/51) * (13/50) * (13/49) = 0.1055

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