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Prob. question

  1. Sep 7, 2003 #1
    This might seem very basicto some of you, I'm just not a math person, and wondering if someone could help me with this:

    A, B, and C shoot at a target in this order: ABCABCABCABC... until the target is hit. Each shot and each shooter is independent of each other. The probability that A hits the target is 4/5. The probability that B hits the target is 4/5. The probability that C hits the target is 1/3. What is the probability that the first person to hit the target is B?

    Could anyone explain how you do this?
  2. jcsd
  3. Sep 7, 2003 #2


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    For any cycle, the event that B hits first is A miss followed by B hit (prob = 4/25). The prob that all three miss on a cycle is 2/75.
    The net result is that the prob that B hits first is
    (4/25)*(1+2/75+(2/75)2+ ...)=(4/25)*(75/73)=12/73.
  4. Sep 7, 2003 #3
    Thanks a lot, but could I ask for the logic behind the (1+2/75+(2/75)^2) part? I'd highly appreciate it.
  5. Sep 8, 2003 #4


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    The probability that everyone misses on a given "cycle" is
    (1/5)(1/5)(2/3)= 2/75. That's the number in the series.

    The "1" is if B hits on the first cycle.
    "2/75" would be where everyone misses on the first cycle, then B hits on the second.
    "(2/75)^2" is where everyone misses on the first two cycles, B hits on the third.
    "(2/75)^n" would be everyone missing on the first n cycles, B hits on the next.

    Of course, that forms a geometric series so it is easy to sum.
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