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Math Help - Probabilty question I'm stuck on

  1. #1
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    Probabilty question I'm stuck on

    Just doing some practice probability questions and came across this which confused me:

    Two darts players
    A and B throw alternately at a board and the
    1st to score a bull wins the game. The outcomes of different throws are independent and on each throw
    A has probability pA of scoring a bull while B has probability pB of scoring a bull. If A has the 1st throw, calculate the probability that A wins the game.


    I'm not sure what type of question this is (though i would guess at geometric). I'm not sure how to work this out as there is no information about the number of throws and I'm also unsure about how to bring independence into this.
    Any advice would be welcome
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  2. #2
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    Quote Originally Posted by schteve View Post
    Just doing some practice probability questions and came across this which confused me:

    Two darts players
    A and B throw alternately at a board and the
    1st to score a bull wins the game. The outcomes of different throws are independent and on each throw
    A has probability pA of scoring a bull while B has probability pB of scoring a bull. If A has the 1st throw, calculate the probability that A wins the game.


    I'm not sure what type of question this is (though i would guess at geometric). I'm not sure how to work this out as there is no information about the number of throws and I'm also unsure about how to bring independence into this.
    Any advice would be welcome
    Yes, schteve,

    it's a geometric series for which you must evaluate S_{\infty}

    A wins if he scores on the 1st throw,
    or misses on the 1st, B misses, A scores on the 2nd,
    or A misses, B misses, A misses, B misses, A scores on the 3rd,
    and so on...........

    The probability is

    P_A+(1-P_A)(1-P_B)P_A+(1-P_A)(1-P_B)(1-P_A)(1-P_B)P_A+...

    =P_A+(1-P_A)(1-P_B)P_A+[(1-P_A)(1-P_B)]^2P_A+[(1-P_A)(1-P_B)]^3P_A+....

    a geometric series with first term P_A and multiplier [(1-P_A)(1-P_B)]

    If you let Q_A=1-P_A,\ Q_B=1-P_B

    it is neater to express.
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