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Probability: what are the chances of shooting a target at least once?

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data

    This is actually from a game that I play online so sorry for the crude question. I have a gun with 3 bullets and 3 targets. The gun randomly shoots the targets. What is the probability that I'll hit target A at least once?

    2. Relevant equations



    3. The attempt at a solution

    P(hitting target A at least once)
    =1 - P(all 3 bullets hitting target A)
    =1 - (1/3)*(1/3)*(1/3)
    =1 - 1/9
    =8/9

    So there is a 8/9 probability of hitting target A at least once. I just wanted to make sure that this is correct.
     
  2. jcsd
  3. Dec 26, 2013 #2

    HallsofIvy

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    I don't understand your logic. Certainly "all three bullets hitting A" would be an example "hitting A at least once". You should have, rather 1 minus the probability of NO bullets hitting A.
    (Also, (1/3)(1/3)(1/3)= 1/27, not 1/9.)
     
  4. Dec 26, 2013 #3
    How about now?

    P(hitting target A at least once)
    =1 - P(no bullets hitting target A)
    =1 - (2/3)*(2/3)*(2/3)
    =1 - 8/27
    =27/27 - 8/27
    =19/27
     
  5. Dec 26, 2013 #4

    haruspex

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    That works.
     
  6. Dec 27, 2013 #5

    Mentallic

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    This game wouldn't happen to be Hearthstone? Of the little I've seen of that game, I have witnessed random shots being taken with certain cards being drawn.

    You can also extend the problem to shooting k times with n targets. Hitting a specific target at least p times where [itex]0\leq p \leq k[/itex] is described by the binomial distribution function, so depending on whether p is closer to 0 or k depends on which following formula you would use:


    [tex]P(\text{hit specific target at least p times})[/tex]
    For p close to 0:
    [tex]= 1-\sum_{i=0}^{p-1}\binom{k}{i}\left(\frac{1}{n}\right)^i\left(1-\frac{1}{n}\right)^{k-i}[/tex]

    and for p close to k:
    [tex]= \sum_{i=p}^{k}\binom{k}{i}\left(\frac{1}{n}\right)^i\left(1-\frac{1}{n}\right)^{k-i}[/tex]
     
  7. Dec 30, 2013 #6
    If you miss all of them, won't it become 4 possibilities?
     
  8. Dec 30, 2013 #7
    @Mentallic: nice, you guessed the right game. Also, it's cool that you generalized it.
     
  9. Dec 30, 2013 #8

    Mentallic

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    I was a big Blizzard fan back in the Diablo 2 days :smile:
     
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