Probability question

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    Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

    Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 4/5 and the probability of Geri hitting the target is always 2/3 , what is the probability that Tom wins the competition?
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    (Original post by staircase)
    If the probability of Tom hitting the target is always 5/4 and the probability of Geri hitting the target is always 3/2
    I do not understand those probabilities. Can you rephrase or type the exact question in full?
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    (Original post by staircase)
    Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

    Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 5/4 and the probability of Geri hitting the target is always 3/2 , what is the probability that Tom wins the competition?

    How can the probabilities be more than 1?
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    (Original post by chazwomaq)
    I do not understand those probabilities. Can you rephrase or type the exact question in full?
    sorry must ave been a copy and past issue, Ill edit it now
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    (Original post by staircase)
    Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

    Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 4/5 and the probability of Geri hitting the target is always 2/3 , what is the probability that Tom wins the competition?
    For Tom to win, he needs to hit and then Geri needs to fail. They could keep
    going forever if they both either keep hitting it or keep failing.

    For Tom to win, he needs to hit = 4/5 AND
    Geri needs to not hit = 1/3.

    4/5 * 1/3 = 4/15. However this alone is not the answer.

    There's also the probability that Tom could win later on.
    Let's say he wins on the 2nd round... the probability is that 1/5*1/3 (both lose) * 4/15 = 1/15 * 4/15 = 4/225.
    or they both win = 4/5*2/3 = 8/15 then *4/15 = 32/225
    Add these two together... 36/225

    I actually don't even know where I'm going with this....... or maybe....

    3rd round = 1/15*^how many times both lost (2 in this case) * 4/15 = 1/225 * 4/15 = 4/3375

    I'm probably wrong, but I think it is ((1/15)^n * 4/15)+((8/15)^y * 4/15) where n is the number of times both have not hit and y is where they've both hit. Someone correct me...
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    (4/5 *1/3) Tom hits other doesn't so he wins route 1 of winning
    (1/5* 1/3) * (4/5*1/3) Both miss first time but Tom hits second time route 2 of winning

    You add these two probabilities up giving your answer ill add the second method too

    (1/5*2/3) Tom misses and Geri hits meaning he doesn't win route 1
    (4/5*2/3) * (1/5*2/3) Both hit first time and then Tom misses and Geri hits meaning he doesn't win route 2

    and so on this actually really long but sometimes it works out better to work out the probability of something not happening and then minus 1. They could also both hit it first time then miss or miss first time and hit or miss miss or hit hit it is really long. Also watch out for questions where it conditional for example if they miss once they have higher chance but that isn't part of the question for now
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    The probability Tom wins in the first round is 4/15.
    The probability the game goes on to a second round is (4/5 * 2/3) + (1/3 * 1/5) = 9/15
    Once they are in the second round it is the same probability of winning as he did before the first round, say p. It is as if the game has reset.
    So p = 4/15 + (9/15)*p.
    Rearranging give (6/15)*p = 4/15
    and the answer is p = 2/3.
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    Is there a limit to how many attempts they can have because this could go on forever?
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    (Original post by 59wlb)
    The probability Tom wins in the first round is 4/15.
    The probability the game goes on to a second round is (4/5 * 2/3) + (1/3 * 1/5) = 9/15
    Once they are in the second round it is the same probability of winning as he did before the first round, say p. It is as if the game has reset.
    So p = 4/15 + (9/15)*p.
    Rearranging give (6/15)*p = 4/15
    and the answer is p = 2/3.
    why do you do (9/15)p??

    PS: I understand 9/15 but why * it by p?
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    (Original post by staircase)
    why do you do (9/15)p??

    PS: I understand 9/15 but why * it by p?
    It uses something called the law of total probability.
    P(A) = P(A|B)*P(B) + P(A|C)*P(C) + P(A|D)*P(D)
    where B, C, and D cover all possible outcomes. (i.e. P(B)+P(C)+P(D)=1. It doesn't have to be three terms, but I am using three for this example).

    In this case A is the probability that Tom wins, B is the probability that he wins in the first round, C is the probability that no one wins in the first round, and D is the probability that Geri wins in the first round.

    Then we have:
    P(B) = 4/15
    P(C) = 9/15
    P(D) = 2/15
    P(A|D) = 0 since Tom can't win if the other person has already won.
    P(A|B) = 1 since he has already won in the first round.
    P(A|C) = P(A) since the first round doesn't affect the second round.

    Then P(A) = 1*4/15 + P(A)*9/15 + 0*2/15 is the equation I had before.
    I hope that's a bit clearer.
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    (Original post by staircase)
    why do you do (9/15)p??

    PS: I understand 9/15 but why * it by p?
    I think I've thought of an easier way of thinking about it, so you might want to ignore my last reply.
    9/15 is the probability the game goes to a second round.
    But we also want Tom to win.
    So we need P(second round AND tom wins)=P(second round)*P(Tom wins)
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    (Original post by staircase)
    Hi, can someone explain to me why the answer to the following question is not simply (4/15)...

    Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 4/5 and the probability of Geri hitting the target is always 2/3 , what is the probability that Tom wins the competition?
    Is this GCSE?
    Have you tried probability tree diagram?
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    Tom wins on the first round with probability 4/5 * 1/3 = 4/15
    Geri wins on the first round with probability 1/5 * 2/3 = 2/15

    Thus Tom is twice as likely to win on the first round as Geri. The same is true on any subsequent round, should they reach it. So p(Tom wins) = 2 x p(Geri wins). Someone has to win. So p(Tom wins) + p(Geri wins) = 1.

    Those two equations give the answer.
 
 
 
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