一題F5 probability問題

用戶107609

2012-03-14 3:31 am

probability

5 cards are randomly drawn from a pack of 52 playing cards one by one without replacement. Find the probability that at least 3 Aces are drawn.

我知可以用
(4C3x48C2) / 52C5 + (4C4x48C1) / 52C5
=0.00175

但如果我想用分數黎計到點做?
點解咁樣計會計唔到?
4X3X2X47X48 / 52X51X50X49X48 + 4X3X2X1X48 / 52X51X50X49X48

回答 (1)

wkho28

2012-03-14 6:40 am

✔ 最佳答案

A and B represent Aces and non-Aces.

(i) 3A and 2B can be:
　AAABB
　AABAB
　ABAAB
　　.
　　.
　BBAAA

The total number of cases is: 5!/(3!x2!)=10

The probability in each case is:
(4x3x2x48x47) / (52x51x50x49x48)

The probability that 3A and 2B is:
10x(4x3x2x48x47) / (52x51x50x49x48)

　or

4C3x48C2 / 52C5
=[(4x3x2)/(3x2x1)x(48x47)/(2x1)]
/ [(52x51x50x49x48)/(5x4x3x2x1)]
=10x(4x3x2x48x47) / (52x51x50x49x48)

(ii) 4A and 1B can be:
　AAAAB
　AAABA
　AABAA
　ABAAA
　BAAAA

The probability in each case is:
(4x3x2x1x48) / (52x51x50x49x48)

The total number of cases is: 5!/4!=5

The probability that 4A and 1B is:
=5x(4x3x2x1x48) / (52x51x50x49x48)

　or

4C4x48C1 / 52C5
=[(4x3X2x1)/(4x3x2x1)x48] / [(52x51x50x49x48)/(5x4x3x2x1)]
=5x(4x3x2x1x48) / (52x51x50x49x48)

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