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Probability C Hits the Target First


Date: 09/08/2001 at 12:18:12
From: rob
Subject: Statistics/probability

A, B and C shoot at a target. A shoots first, followed by B, then by 
C. They keep shooting in this sequence: A,B,C,A,B,C... until the 
target is hit. A hits the target 5/6 of the time, B hits the target 
3/4 of the time, and C hits the target 2/3 of the time. Find the 
probability that C is the first one to hit the target.

Once I understand how to do this problem I'm sure others will make 
sense to me; I just don't know where to start and what to do.
Thank you so much!

Rob


Date: 09/08/2001 at 13:35:50
From: Doctor Mitteldorf
Subject: Re: Statistics/probability

Dear Rob,

This is indeed a tricky problem - but your experience with algebra 
will serve you. I'm going to tell you what to do, then I'll let you 
actually do the work. Of course, if you need more hints, or if you get 
an answer you're not sure of, feel free to write back and I'll try to 
respond as quickly as I can...

Here's what to do: 

Analyze the first cycle exhaustively, with all the probabilities. The 
probability that A misses is 1/6. Multiply that by the probability 
that B misses, and multiply the product by the probability that C HITS 
the target. Now you have the probability that C will be the first
one to hit the target IN THE FIRST CYCLE.  

The next thing you might think of is: Now calculate the second and 
third and fourth cycles the same way. Surely we'll see a pattern, and 
the probabilities are getting smaller rapidly. This process must 
converge, and maybe we'll get an infinite series that I know how to 
sum...

Well, that's right - and that way would work, but here's a shortcut. 
Just above, we calculated the probability that C would win in the 
first cycle. That's a number that you know. We'll call that number c.  
Now let p be the probability that C wins in ANY cycle. We'll be really 
clever and write an equation that connects p to c. Here's how:

p can be broken down into two parts. One is the probability that C 
wins in the first cycle, which is c. So p = c + something. The second 
part is the probability that he wins in cycle 2, 3, 4, or any other 
cycle. Here's the trick: if C fails to win the first cycle, all three 
contestants are exactly back where they started. So we can write an 
expression for the second part of p: it is (1-c) times p. In other 
words, the equation you will write says, either C wins in the first 
round, with a probability c which we know, or else C loses in the 
first round, with a probability (1-c), and then we're back where we 
started from and, by definition, his probability of winning is p.

In this way, you'll construct an equation - not a formula for p, since 
p will appear on both sides, but a linear equation for p which you can 
solve easily enough with elementary algebra.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Probability

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