標題:
maths
發問:
In a deck of 52 cards,Amy gets 5cards randomly from it.(a) Find the probability that she gets one pair of''A'''.(b)Find the probability that she gets 2 pairs.(c)Find the probability that she gets 3''A'' and 1 pair.(d)Find the probabilty that she gets 4 of a same... 顯示更多 In a deck of 52 cards,Amy gets 5cards randomly from it. (a) Find the probability that she gets one pair of''A'''. (b)Find the probability that she gets 2 pairs. (c)Find the probability that she gets 3''A'' and 1 pair. (d)Find the probabilty that she gets 4 of a same kind.
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最佳解答:
(a) No. of ways to get 5 cards randomly from 52 cards = 52C5 To get one pair of "A" : ? Get 2 "A" out of the 4"A" (4C2). ? Then get 3 cards from the 48 cards that are NOT "A" (48C3). The required probability = 4C2 x 48C3 / 52C5 = (4! / 2!2!) x (48! / 3!45!) / (52! / 5!47!) = 2162 / 54145 ===== (b) To get two pairs : ? Choose 2 kinds from the 13 kinds (13C2). ? Get 2 cards from the 4 cards of each of the 2 kinds ((4C2)2)chosen above. ? Get 1 cards from the 44 cards of the other 11 kinds (44C1). The required probability = 13C2 x (4C2)2 x 44C1/ 52C5 = (13! / 2!11!) x (4! / 2!2!)2 x (44! / 1!43!) / (52! / 5!47!) = 198 / 4165 ===== (c) To get 3"A" and 1 pair : ? Get 3 "A" out of the 4 "A" (4C3). ? Choose 1 kind from the 12 kinds apart from "A" (12C1). ? Get 2 cards from the 4 cards of the kind chosen (4C2). The required probability = 4C3 x 12C1 x 4C2/ 52C5 = (4! / 3!1!) x (12! / 1!11!) x (4! / 2!2!) / (52! / 5!47!) = 6 / 54145 ===== (d) To get 4 of a same kind : ? To choose 1 kind from the 13 kinds (13C1). ? Get all 4 cards of the kind chosen (1). ? Get 1 card from the 48 cards of the rest 12 kinds (48C1). The required probability = 13C1 x 48C1 / 52C5 = (13! / 1!12!) x (48! / 1!47!) / (52! / 5!47!) = 1 / 4165
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