1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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


    User Avatar
    Science Advisor

    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
  5. Dec 26, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That works.
  6. Dec 27, 2013 #5


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    I was a big Blizzard fan back in the Diablo 2 days :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted