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Probabilities of archers A and B hitting the bull's eye

  1. Jan 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Two archers A and B take turns to shoot, with archer A taking the first shot. The probabilities of archers A and B hitting the bull's eye is 1/6 and 1/5 respectively. Show that the probability of archer A hitting the bull's eye first is 1/2.


    2. Relevant equations



    3. The attempt at a solution
    Well, i thought about drawing tree diagram but the problem is I have no idea when to stop because I don't know when he will hit.
     
  2. jcsd
  3. Jan 10, 2013 #2

    dx

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    Re: Probability

    First, what are the possible ways for this to happen?

    A could hit the bulls eye on his first attempt. Or, A misses, B misses and then A hits the bulls eye. Or, A misses, B misses, A misses, B misses and then A hits the bulls eye... and so on.

    Find the probability for each of these ways (for A to hit the bull's eye first), and add them up.

    You will need to use the following:

    1 + a2 + a3 + ... = 1/(1-a) for |a| < 1
     
  4. Jan 10, 2013 #3
    Re: Probability

    the problem is, if he hits on the first attempt that's 1/6. so If i go on, it will be 1/6 + ( 5/6 * 4/5 * 5/6 * 4/5 * 5/6 * 4/5....) I have no idea when to stop. Moreover, the product of probability he misses will be come smaller and smaller and that number + 1/6 will not give me half i think.
     
  5. Jan 10, 2013 #4

    dx

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    Re: Probability

    Let's take the case when A gets it on his second try: A misses, B misses, and then A hits. What is the probability for this?
     
  6. Jan 10, 2013 #5
    Re: Probability

    that will be 5/6 * 4/5 * 1/6 right?
     
  7. Jan 10, 2013 #6
    Re: Probability

    I'm sorry but what's this?
    '1 + a2 + a3 + ... = 1/(1-a) for |a| < 1'
     
  8. Jan 10, 2013 #7

    dx

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    Re: Probability

    Yes, correct.

    Now what is the probability that A gets it on his 3rd try? (A misses, B misses, A misses, B misses, A hits bull's eye)

    Ignore that for the moment. We'll get to it soon.
     
  9. Jan 10, 2013 #8
    Re: Probability

    5/6* 4/5 * 5/6 * 4/5 * 1/6
     
  10. Jan 10, 2013 #9

    dx

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    Re: Probability

    Correct.

    So do you see the pattern?

    First try: 1/6

    Second try: (1/6)(4/6)

    Third try: (1/6)(4/6)2

    Fourth try: (1/6)(4/6)3

    ...

    What is the probability that he gets in on the n'th try? Each time you are simply multiplying the previous one by (5/6)(4/5) = 4/6
     
  11. Jan 10, 2013 #10
    Re: Probability

    1/6 * 4/6 ^ n-1!
     
  12. Jan 10, 2013 #11

    dx

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    Re: Probability

    Exactly.

    Now you simply add them up. Let a = 4/6

    (1/6) + (1/6)a + (1/6)a2 + (1/6)a3...

    which is (1/6)(1 + a + a2 + a3 + a4...)

    This is where you use the formula I mentioned before.
     
  13. Jan 10, 2013 #12
    Re: Probability

    Got it! Thanks alot!!:smile:
     
  14. Jan 10, 2013 #13

    dx

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    Re: Probability

    No problem.
     
  15. Jan 10, 2013 #14

    Ray Vickson

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    Re: Probability

    Here is another approach: let x = the probability that A will win, given that A is just about to shoot.

    If A hits the target on his first shot, he wins; the probability of this is 1/6. If A misses his first shot (prob. = 5/6), then for A to win it must be the case that B misses his first shot (prob = 4/5); at that point, A is just about to shoot and we are back to the starting point (win prob. = x at that point). So, altogether we have
    [tex]x = \frac{1}{6} + \frac{5}{6} \frac{4}{5}\, x = \frac{1}{6} + \frac{2}{3} x.[/tex]
    Solving, we get x = 1/2.
     
  16. Jan 10, 2013 #15

    CAF123

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    Re: Probability

    @Ray
    Is there a reason why doing: $$x = \frac{1}{6} + \frac{2}{3}x + \frac{5}{6} \frac{4}{5} \frac{5}{6} \frac{4}{5}x,$$ and solving is incorrect? (I am just considering the possibility of both A and B losing until A hits on his third go).
     
  17. Jan 10, 2013 #16
    Re: Probability

    My friend Kim and I had a similar competition, but we're much worse. I am slightly worse than Kim, so I went first. On any one shot, my chance of hitting the bull is 1/39, and Kim's chance is 1/38.

    Show that, once again, this is an even competition - my chance of getting the bull's eye first is 1/2.
     
  18. Jan 10, 2013 #17

    CAF123

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    Re: Probability

    Are you stuck? If so, just use exactly the same method as you used to solve the first one. Alternatively, say you win on your ##i##th shot. This means you (or Kim) cannot have hit the bull on the previous ##i-1## attempts.
    What is the prob of you not hitting the bull on the ##i-1## attempts?
    What is the prob of Kim not hitting the bull on her ##i-1## attempts?
    What is the prob of you hitting on the ##i##th trial given that you and Kim failed on the ##i-1## trials?
    By the independence of trials you can then multiply these together and sum.
    (This is really the same method by dx just reworded a bit differently)
     
  19. Jan 10, 2013 #18

    haruspex

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    Re: Probability

    Just use Ray's method again. It's much the simplest way to solve these problems. Did you understand Ray's method?
     
  20. Jan 10, 2013 #19
    Re: Probability

    Sorry for the misunderstanding guys. I was giving an extension question to illustrate the reasoning; I'm not stuck.
     
  21. Jan 10, 2013 #20

    Ray Vickson

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    Re: Probability

    We can see it is incorrect by solving it and getting the wrong answer (x = -3/2 instead of x = 1/2). To understand WHY it is wrong, go back and look at the definition of x and the argument I gave for the equation satisfied by x. The point is that x already includes all the possibilities of winning on the first, third, fifth, .... go.

    However, it would be correct to write
    [tex]x = \frac{1}{6} + \frac{2}{3}\frac{1}{6} + \frac{2}{3}\frac{2}{3}\,x [/tex]
    for similar reasons. This equation does have the correct solution x = 1/2.
     
    Last edited: Jan 10, 2013
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