Probability C Hits the Target FirstDate: 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/ |
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