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Probability homework problem

  1. Jul 19, 2008 #1
    1. The problem statement, all variables and given/known data
    Two shooters aim in same target at same time and then the first shooter shots in 70% of the shooting session, and the second is missing in the 60% of the shooting sesion.

    A. The probability to miss the first shooter is: ______________

    B. The probability the target to be shot is: _______________


    2. Relevant equations


    3. The attempt at a solution

    A. I think it is 30% (very easy indeed)

    So the first shooter shots 70% (7 of 10) and the second shooter 40% (4 of 10)

    B. I think the probability is more than 70%, but what is the correct one?

    Thank you.
     
  2. jcsd
  3. Jul 19, 2008 #2

    Dick

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    Re: probability

    If you found A easy then what's the probability BOTH shooters miss? What's the relation of that probability to the probability that the target is hit?
     
  4. Jul 20, 2008 #3
    Re: probability

    The both shooters to miss is probably 55% (or 5,5 of 10)

    [tex]1-\frac{5,5}{10}=\frac{4,5}{10}[/tex]

    I don't think this is correct.
     
  5. Jul 20, 2008 #4

    Dick

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    Re: probability

    You're right. It's not. Odds A misses are 0.30, odds B misses are 0.60. Odd that they both miss is NOT the average of the two. What is it?
     
  6. Jul 20, 2008 #5
    Re: probability

    If the first one misses 3 of 10, and the second one 6 of 10, both miss 9 of 20, or 45%. I don't really know. Please help!
     
  7. Jul 20, 2008 #6

    HallsofIvy

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    Re: probability

    Do you know how to find P(A and B) if you know P(A), P(B), and know that A and B are independent?
     
  8. Jul 20, 2008 #7
    Re: probability

    Ok, I will do like this:
    1-shot 2-miss

    shooter C - 1111111222

    shooter D - 1111222222

    All possible combinations are:
    7*(1,1 ; 1,1 ; 1,1 ; 1,1 ; 1,2 ; 1,2 ; 1,2 ; 1,2; 1,2; 1,2)+3*(2,1 ; 2,1; 2,1 ; 2,1 ; 2,2 ; 2,2; 2,2 ; 2,2 ;2,2 ;2,2)
    10*7+3*10=70+30=100

    [tex]\frac{10*7+4*3}{100}=\frac{70+12}{100}=\frac{82}{100}[/tex]

    82% ?
     
  9. Jul 20, 2008 #8

    Redbelly98

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    Re: probability

    82% is correct.
     
  10. Jul 21, 2008 #9
    Re: probability

    Ok, thanks. But how will I solve it with permutations?
     
  11. Jul 21, 2008 #10
    Re: probability

    Some help please?
     
  12. Jul 21, 2008 #11
    Re: probability

    for second question ::

    probability that A miss :p(A): 30/100
    probability that B miss :p(B): 60/100

    Probability that both Miss :: p(AnB) = p(A) x p(B) = 30/100 * 60/100 = 18/100 (As both Events are independent)

    so the Probability that the Target is shot = 1 - Both Miss
    1 - 18/100 = 82/100 = 82 %.
     
  13. Jul 21, 2008 #12
    Re: probability

    Where this formula comes from?
     
  14. Jul 21, 2008 #13
  15. Jul 22, 2008 #14
    Re: probability

    Ok, thank you.
     
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