匿名
匿名 發問於 科學及數學數學 · 7 月前

Probability(急)?

題目: A bag contains 4 yellow balls, 2 green balls and 1 red ball. In a game, two players A and B take a ball alternately from the bag with replacement. The first player who takes the red ball wins the game. If player A takes a ball first, what is the probability that player A will win the game.

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  • 匿名
    7 月前
    最愛解答

    Question:

     A bag contains 4 yellow balls, 2 green balls and 1 red ball. In a game, two players A and B take a ball alternately from the bag with replacement. The first player who takes the red ball wins the game. If player A takes a ball first, what is the probability that player A will win the game.

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    Solution:

    The probability that the player takes a red ball in the first chance

    = 1/(4 + 2 + 1)

    = 1/7

    The probability that the player takes a red ball in the second chance

    = (1 - 1/7) (1 - 1/7) (1/7)

    = (6/7)² (1/7)

    The probability that the player takes a red ball in the third chance

    = (1 - 1/7) (1 - 1/7) (1 - 1/7) (1 - 1/7) (1/7)

    = (6/7)⁴ (1/7)

    The probability that player A will win the game

    = 1/7 + (6/7)² (1/7) + (6/7)⁴ (1/7) + (6/7)⁶ (1/7) + ...

    = (1/7) / [1 - (6/7)²]

    = 7/13

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    Remark:

    The sum to infinity of a geometric series = a/(1 - r) where

    a is the first term,

    r is the common ratio and -1 < r < 1.

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