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Math Help - Probability - shooting

  1. #1
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    Probability - shooting

    Two guys are shooting at a target. First one fires 9 rounds and second one fires 10 rounds in this session. First one has 8/10 precision and second one has 7/10 precision. There is a hit. What is the probability that second shooter is the one that hit it?
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  2. #2
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    Quote Originally Posted by losm1 View Post
    Two guys are shooting at a target. First one fires 9 rounds and second one fires 10 rounds in this session. First one has 8/10 precision and second one has 7/10 precision. There is a hit. What is the probability that second shooter is the one that hit it?
    From a tree diagram and Bayes Theorem:

    \Pr(\text{ Second shooter} | \text{ Hit}) = \frac{\frac{10}{19} \cdot \frac{7}{10}}{\frac{9}{19} \cdot \frac{8}{10} + \frac{10}{19} \cdot \frac{7}{10}} = ....
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  3. #3
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    I'm a bit confused.

    Bayes' Theorem is:

    Pr(a|b)=\frac{Pr(b|a)*Pr(b)}{((Pr(b|a)*Pr(a))+((Pr  (b|a^c)*(Pr(a^c))}

    How did you come to find that Pr(b|a) is equal to 10/19?
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  4. #4
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    I'm a bit confused.

    Bayes' Theorem is:



    How did you come to find that Pr(b|a) is equal to 10/19?
    there is a hit
    Rounds fired by both the guys = 19

    Rounds fired by Second one = 10

    Probability that the round that hit the target was fired by 2nd guy = 10/19

    Probability that the second one hit the target = 7/10

    This should occur simultaneously

    ......What's wrong with that ?I don't find anything wrong in MrF's reply
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  5. #5
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    Quote Originally Posted by downthesun01 View Post
    I'm a bit confused.

    Bayes' Theorem is:

    Pr(a|b)=\frac{Pr(b|a)*Pr(b)}{((Pr(b|a)*Pr(a))+((Pr  (b|a^c)*(Pr(a^c))}

    How did you come to find that Pr(b|a) is equal to 10/19?
    Did you draw the tree diagram?
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  6. #6
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    Quote Originally Posted by losm1 View Post
    Two guys are shooting at a target. First one fires 9 rounds and second one fires 10 rounds in this session. First one has 8/10 precision and second one has 7/10 precision. There is a hit. What is the probability that second shooter is the one that hit it?
    Looking at it logically, we get

    the 2nd shooter hits the target an average of 7 out of 10 times,
    so it's expected he should hit the target 7 times from his 10 attempts
    (or 70 times per hundred etc)

    the first shooter hits the target an average of 8 times out of 10,
    so he ought to hit the target an average of 9(0.8)=7.2 per 9,
    or 72 times per ninety etc.

    Then, the probability of a single hit belonging to one or the other is

    \frac{(number\ of\ A\ hits)}{(number\ of\ A\ hits)+(number\ of\ B\ hits)}=\frac{70}{70+72} for the 2nd shooter.
    Last edited by Archie Meade; April 22nd 2010 at 04:52 PM. Reason: typo
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