Prob. question

  • Thread starter guevara
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  • #1
guevara

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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?
 

Answers and Replies

  • #2
mathman
Science Advisor
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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.
 
  • #3
guevara
Thanks a lot, but could I ask for the logic behind the (1+2/75+(2/75)^2) part? I'd highly appreciate it.
 
  • #4
HallsofIvy
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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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