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  1. #1
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    Exclamation Probabilty Help urgent help

    [LEFT]Two important members of a cricket team are injured, and each has probability 1/3 of recovering before the match. The recoveries of the two players are independent of each other. If both are able to play then the team has probability 3/4 of winning the match, if only one of them plays then the probability of winning is 1/2 and if neither play the probability of winning is 1/16. What is the probability that the match is won?
    Last edited by maths_2468; November 15th 2008 at 05:51 AM.
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  2. #2
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    Hi there.

    Quote Originally Posted by maths_2468 View Post
    [left]Two important members of a cricket team are injured, and each has probability 1/3 of recovering before the match. The recoveries of the two players are independent of each other. If both are able to play then the team has probability 3/4 of winning the match, if only one of them plays then the probability of winning is 1/2 and if neither play the probability of winning is 1/16. What is the probability that the match is won?
    probability(no one recovers and the game is won) = 2/3 * 2/3 *1/16

    prob(one guy recovers and the game is won) = 2/3*1/3 * 1/2 + 1/3 * 2/3 *1/2

    prob(everyone recovers and the game is won) = 2/3*2/3 * 3/4

    prob("game is won")=2/3 * 2/3 *1/16 + 2/3*1/3 * 1/2 + 1/3 * 2/3 *1/2 + 2/3*2/3 * 3/4

    i just added all three probabilities and i got prob() = 0.5833....3
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  3. #3
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    Quote Originally Posted by Rapha View Post
    probability(no one recovers and the game is won) = 2/3 * 2/3 *1/16
    prob(one guy recovers and the game is won) = 2/3*1/3 * 1/2 + 1/3 * 2/3 *1/2

    prob(everyone recovers and the game is won) = 2/3*2/3 * 3/4
    There are two mistakes in the above. It should be:
    prob(everyone recovers and the game is won) = 1/3*1/3 * 3/4
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