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發問:
Cards are drawn at random from an ordinary deck of 52, one by one and without replacement. What is the probability that no Heart is drawn before the Ace of Spades?
最佳解答:
you could only consider 14 cards out of 14 so just imagine there are only 14 cards, 13 hearts and 1 Ace of Spades so it is easy to find that the probability that no Heart is drawn before the Ace of Spades is 1/14(the Ace of Spades is the first one out of 14) any problem? 2007-02-07 23:44:29 補充： correction:you could only consider 14 cards out of 52 2007-02-08 01:22:15 補充： 只要明白呢條係比較14隻牌o既先後次序而唔係比較52隻牌, 就好易搵到答案.
其他解答:
By means of permutation: For 52 cards, there are 52! possibilities of arrangement of the cards where ! means the factorial operation. Now, let A be the card of ace of spades and then take a look at the diagram below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf.jpg 【此圖乃本人自製圖片，未經本人同意勿擅自連結或使用】 In fact, card A can be from position 1 to 39 for any favourable outcome (since at least 13 positiosn should be left at the end for those hearts), i.e. 1 ≦ r ≦ 39 Now, for each value of r, if we want a favourable outcome, the preceding r-1 positions before A must be filled by non-heart and for the remaining 52 - r position behind A, the arrangement can be random. So the general expression of number of permutations for any 1 ≦ r ≦ 39 is: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf2.jpg And hence the final probability will be summing up all these values for 1 ≦ r ≦ 39 divided by the total number of possible permutations, i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf3.jpg ∴ The probability, if expressed in percentage, is about 7.14%.
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probability發問:
Cards are drawn at random from an ordinary deck of 52, one by one and without replacement. What is the probability that no Heart is drawn before the Ace of Spades?
最佳解答:
you could only consider 14 cards out of 14 so just imagine there are only 14 cards, 13 hearts and 1 Ace of Spades so it is easy to find that the probability that no Heart is drawn before the Ace of Spades is 1/14(the Ace of Spades is the first one out of 14) any problem? 2007-02-07 23:44:29 補充： correction:you could only consider 14 cards out of 52 2007-02-08 01:22:15 補充： 只要明白呢條係比較14隻牌o既先後次序而唔係比較52隻牌, 就好易搵到答案.
其他解答:
By means of permutation: For 52 cards, there are 52! possibilities of arrangement of the cards where ! means the factorial operation. Now, let A be the card of ace of spades and then take a look at the diagram below: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf.jpg 【此圖乃本人自製圖片，未經本人同意勿擅自連結或使用】 In fact, card A can be from position 1 to 39 for any favourable outcome (since at least 13 positiosn should be left at the end for those hearts), i.e. 1 ≦ r ≦ 39 Now, for each value of r, if we want a favourable outcome, the preceding r-1 positions before A must be filled by non-heart and for the remaining 52 - r position behind A, the arrangement can be random. So the general expression of number of permutations for any 1 ≦ r ≦ 39 is: 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf2.jpg And hence the final probability will be summing up all these values for 1 ≦ r ≦ 39 divided by the total number of possible permutations, i.e. 圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Probphlmf3.jpg ∴ The probability, if expressed in percentage, is about 7.14%.
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