Probability

2015-02-01 2:07 am
1.Three cards are drawn from an ordinary 52-card deck without replacement
What is the probability that the first two cards are aces and the third card is a king?Assume that at each step, each one of the remaining cards is equally likely to be
picked.

回答 (3)

2015-02-01 2:53 am
✔ 最佳答案
Pr(1st ace and 2nd ace and 3rd king)

= Pr(1st ace) * Pr(2nd ace and 3rd king | 1st ace)

= Pr(1st ace) * Pr(2nd ace | 1st ace) * Pr(3rd king | 1st ace and 2nd ace)

= (4/52) * (3/51) * (4/50)

= (1/13) * (1/17) * (2/25)

= 2/5525


2015-02-01 17:30:12 補充:
嗚謝 sioieng 提供另解:

The required probability

= Pr(2 aces from 52 cards with 4 aces) × Pr(a king from 50 cards with 4 kings)

= [C(4,2) / C(52,2)] × [C(4, 1) / C(50, 1)]

= [(4!/2!2!) / (52!/50!2!)] × [(4!/1!3!) / (50!/1!49!)]

= [(4 × 3) / (52 × 51)] × [4 / 50]

= [(1 × 1) / (13 × 17)] × [2 / 25]

2015-02-01 17:30:28 補充:
= 2/5525

2015-02-01 17:30:52 補充:
嗚謝 土扁 師兄 提供另解:

The required probability

= Pr(1 ace from 52 cards with 4 aces) × Pr(1 ace from 51 cards with 3 aces) × Pr(a king from 50 cards with 4 kings)

= (4/52) × (3/51) × (4/50)

= (1/13) × (1/17) × (2/25)

= 2/5525
2015-02-02 1:11 am
The required probability

= Pr(1 ace from 52 cards with 4 aces) × Pr(1 ace from 51 cards with 3 aces) × Pr(a king from 50 cards with 4 kings)

= (4/52) × (3/51) × (4/50)

= (1/13) × (1/17) × (2/25)

= 2/5525
2015-02-02 1:07 am
The required probability

= Pr(2 aces from 52 cards with 4 aces) × Pr(a king from 50 cards with 4 kings)

= [C(4,2) / C(52,2)] × [C(4, 1) / C(50, 1)]

= [(4!/2!2!) / (52!/50!2!)] × [(4!/1!3!) / (50!/1!49!)]

= [(4 × 3) / (52 × 51)] × [4 / 50]

= [(1 × 1) / (13 × 17)] × [2 / 25]

= 2/5525


收錄日期: 2021-04-15 18:06:39
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