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.