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Probability: B needs to shoot more times than A

  1. Jun 17, 2010 #1
    Hi

    I am having some problems solving this exercise. Can somebody give a hint on how to solve this. The hint from the book is not really helping me.

    1. The problem statement, all variables and given/known data

    Two sharpshooters, A and B, are going to shoot at a target. A has probability Pa of hitting it on a single shot; B has probability Pb of hitting it on a single shot. Wheater the target is struck on any one shot is statistically independent of whether it is struck on any other shot. What is the probability that B needs to shoot more times before hitting the target than A?

    (Hint from the book:
    1. Suppose that A hits the target for the first time on his nth shot.
    2. Calculate the probabity that B shoots more than n times before hitting the target
    3. Then use the principle of total probabity to account for all values of n from 1 ad infinium)

    2. Relevant equations

    Pr(M AND M) = Pr(M|Ai)*Pr(Ai)
    Pr(M) = Pr(M|A1)*Pr(A1) + Pr(M|A2)*Pr(A2) ...

    3. The attempt at a solution

    P(A nth shot) = Pa*(1-Pa)^(n-1)
    P(B nth+1 shot) = Pb*(1-Pb)^n
    ????

    Thank you
     
  2. jcsd
  3. Jun 17, 2010 #2

    vela

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    You could think of it this way. In each trial, each shooter keeps shooting until he hits the target. Say A hits the target on shot nA and B hits the target on shot nB. You have a bunch of possible outcomes (nA,nB). The outcome (1,1) would mean they both hit on their first shot; (1,2) would mean A hits on the first shot and B hits on his second shot; (2,1) would mean A needed two shots and B only one; and so on. What's the probability P(nA,nB) of outcome (nA,nB)?

    Finally, the event you're interested in consists of the outcomes where nB>nA. How would you express that in terms of the probabilities of the individual outcomes?
     
  4. Jun 20, 2010 #3
    > What's the probability [tex]P(n_A,n_B)[/tex] of outcome [tex](n_A,n_B)[/tex]?

    [tex]P(n_a,n_b) = P_a*(1-P_a)^{(n_a-1)}*P_b*(1-P_b)^{(n_b-1)}[/tex]

    > Finally, the event you're interested in consists of the outcomes where [tex]n_B>n_A[/tex].
    > How would you express that in terms of the probabilities of the individual outcomes?

    I have no clue.
     
  5. Jun 20, 2010 #4

    vela

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    Then you better go back and review basic probability.
     
  6. Jun 20, 2010 #5
    > Then you better go back and review basic probability.

    I already did. I guess here is some point i don't understand.
    So is the first assumption i did for [tex]P(n_a,n_b)[/tex] also wrong?
     
  7. Jun 20, 2010 #6

    vela

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    Your expression for P(nA,nB) is correct.
     
  8. Jun 23, 2010 #7
    with the help of
    [tex]\sum_{n=0}^{\infty} q^n = \frac{1}{1-q}[/tex]

    [tex]P(b > a) = \sum_{n_a}^{\infty} \sum_{n_b=n_a+1}^{\infty} P(a_n,a_b) = 1-P_b[/tex]

    is this correct ?
     
  9. Jun 23, 2010 #8

    vela

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    No, it's not correct. Your answer should depend on pA somehow. Let pB be fixed. If pA>pB, you'd expect A would typically hit the target before B does. If pA<pB, you'd expect B would hit the target before A does generally. So the probability P(nB>nA) should be different in the two cases, but your answer would say they're the same.
     
  10. Jun 23, 2010 #9
    Oh i think i found the mistake!

    [tex]\frac{P_a (1-P_b)}{1-(1-P_b)(1-P_a)}[/tex]

    Which looks right to me, because when Pa goes higher, P(nb>na) -> Pb
     
  11. Jun 23, 2010 #10

    vela

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    Looks good!

    The summation could be written

    [tex]P(N_A>N_B) = \sum_{n_a=1}^\infty \sum_{n_b=n_a+1}^\infty P(n_a,n_b) = \sum_{n_a=1}^\infty P_a(1-P_a)^{n_a-1} \sum_{n_b=n_a+1}^\infty P_b(1-P_b)^{n_b-1}[/tex]

    The sum

    [tex]\sum_{n_b=n_a+1}^\infty P_b(1-P_b)^{n_b-1}[/tex]

    is the probability P(NB>NA|N_a=nA) that B shoots more times than A does given that A hits on shot nA. This is the probability that hint 2 asked you to find. The other sum corresponds to hint 3, so you have in the end

    [tex]P(N_A>N_B) = \sum_{n_a=1}^\infty P(N_B>N_A|N_a=n_A)P(N_A=n_A)[/tex]
     
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