Mau Fai 發問於 科學及數學數學 · 1 十年前

are there any easy ways for this question?!!(probability)

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 is drawn?

Let A be ace of spade

my way is :pro(1st drawn is A)+pro(no heart appears in 1st drawn and the 2nd drawn is A)...+pro(no heart appears in the first 38 places and 39 th is A)

that is too complicated, are there anyone who can work it out easily or in another ways?thanks!

4 個解答

評分
  • 1 十年前
    最愛解答

    No other way, I think.

    There's another way of interpreting it, but it gives the same complicated result.

    Place the 52 cards in sequence of the time of being drawn.

    No. of permutations that

    the first (r-1) cards have no heart AND the r-th card is space ace

     = (52-13-1)P(r-1) x 1 x (52-(r-1)-1)! , where nPr = n! / (n-r)!

     = 38P(r-1) x 1 x (52-r)!

     = [ 38! / (39-r)! ] x (52-r)!

    Total no. of permutations that satisfy the requirement

     = sum of "[ 38! / (39-r)! ] x (52-r)!", r from 1 to 39

    Therefore,

    probability

     = { sum of "[ 38! / (39-r)! ] x (52-r)!", r from 1 to 39 } / 52!

  • 1 十年前

    Let me try to explain too. By means of permutation:

    For 52 cards, there are 52! possibilities of arrangement of the cards where ! means the factorial operation.

    Then take a look at the diagram below:

    圖片參考:http://i117.photobucket.com/albums/o61/billy_hywun...

    【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_hywun...

    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_hywun...

    ∴ The probability, if expressed in percentage, is about 7.14%.

    資料來源: My Maths knowledge
  • 1 十年前

    咁, 如果問題太複雜, 首先, 就用一個細的既問題黎諗, 先揾到個pattern 先.

    Q: 有三張牌, 紅心H, 黑葵S, 梅花C. 兜亂, 問 紅心H 係 黑葵S 後面既機會.

    List all combination

    1 HSC

    2 HCS

    3 CHS

    4 CSH

    5 SHC

    6 SCH

    456 H 係S 後面, 所以係 1/2.

    1/2 既意義係... 如果只有兩張牌, S 同 H . H 係 S 後面既機會係幾多? 1/2.

    其實, 第三張牌 C, 係呢個問題係無關架. 你睇吓 3同4, 1同5, 2同6, 呢三對case, 每一對 C 既位置都係一樣. 只係H同S 唔同. 所以34, 15, 26. 每一對case 入面 H 係S 後面都係 1/2. 如果你要 伸長黎寫, 正確既答案 就係 1/2 * 1/3 + 1/2 * 1/3 + 1/2 * 1/3, 所以都係 1/2. 個3 次既 1/3. 其實係多餘既.

    所以你個提既答案係只要考慮 14張牌, 13H + 1S. Spade 係最頭既機會, 其餘個26張牌無關架.

    2007-02-03 14:33:10 補充:

    http://hk.knowledge.yahoo.com/question/?qid=700702... 你揀個答案係錯架, 連題目都解錯, 你問下老師啦, 你快的收番個答案啦.

    2007-02-03 18:17:59 補充:

    http://hk.knowledge.yahoo.com/question/?qid=700702... 是我搞錯, 你選的答案是對的.

  • vsloy
    Lv 7
    1 十年前

    I would like to suggest to slove the question as follow:

    For each card, it may be drawn before or after the Ace of Spade at equal chance.

    The probability that one card is drawn after the Ace of Spade is 1/2.

    No heart is drawn before the Ace of Spade,

    that meant all the 13 cards in heart were drawn after the Ace of Spade.

    Therefore, the probability will be (1/2)^13.

    Please point out if there is any mistake in my suggestion.

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