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Interesting thing about archers hitting the target

  1. Jun 13, 2009 #1
    Hello, again!

    I got one very interesting question.

    We got three archers, and the probability of the ones to hit the target is:
    A1, A2, A3.

    What if the task is to find the probability that the target will be hit at least from one archer.

    So at least one archer to hit the target.

    Is it P(A1 U A2 U A3) = A1 + A2 +A3 ?

    Or [tex](1- P(\bar{A_{1}} \cap \bar{A_{2}} \cap \bar{A_{3}})) = 1 - \bar{A_{1}}* \bar{A_{2}} * \bar{A_{3}}[/tex], where [tex]\bar{A}[/tex] is opposite of A?

    Or maybe, both are valid?

    Thanks in advance.
  2. jcsd
  3. Jun 13, 2009 #2


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    No. This is wrong. [itex]P(A_1\cup A_2)= P(A_1)+ P(A_2)- P(A_1\cap A_2)[/itex] and that extends to 3 events: [itex]P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)[/itex][itex]- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)[/itex].

    Yes, this is correct.

  4. Jun 13, 2009 #3
    Thanks for the post, HallsofIvy.

    In this case, the shootings of the archers are independent cases. So that's why we do not need [itex]P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)- P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)[/itex]
    since [itex]P(A_1\cap A_2)- P(A_1\cap A_3)- P(A_2\cap A_3)+ P(A_1\cap A_2\cap A_3)=0-0-0-0=0[/itex]. That's why I said independent cases.

    But what if P(A1)=0.8, P(A2)=0.9, P(A3)=0.75

    In that case the sum [itex]P(A_1\cup A_2\cup A_3)= P(A_1)+ P(A_2)+ P(A_3)=0.8+0.9+0.75=2.45[/itex]

    This is strange.
  5. Jun 13, 2009 #4
    The fact that two cases A and B are independent doesn't mean P(A n B) = 0. It means that P(A n B) = P(A)P(B).
  6. Jun 13, 2009 #5
    If two cases are independent, that means that they do not have something in common, right? So A n B = 0, right?
  7. Jun 13, 2009 #6


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    A \cap B = \emptyset

    if the two sets are disjoint , which is not the same as independent.
  8. Jun 13, 2009 #7
    Actually, that would mean that they're not independent. Two events are independent if the occurrence of one does not influence the occurrence of the other, i.e. P(A|B) = P(A). If [tex]A \cap B = \emptyset[/tex], then P(A|B) = 0, so the events are not independent.

    With regards to the original question, it would be easiest to take the probability that every archer misses and subtract it from one.
  9. Jun 14, 2009 #8
    Thanks for the replies.
    @Tibarn, in this case the occurrence of one does not influence the occurrence of the other.
    [tex]P(A/B)=\frac{m_{A\cap B} }{ m_{B}}[/tex]
    out of there
    [tex]P(A/B)=\frac{\frac{m_{A\cap B}}{n}}{\frac{m_{B}}{n}}[/tex]
    [tex]P(A/B)=\frac{P(A\cap B)}{ P(B)}[/tex]

    The cases are independent if the occurrence of one does not influence the occurrence of the other. So, if two cases are independent, then P(A/B)=P(A). Out of there P(A n B)=P(A)*P(B)

    Now, let's get back to the task. I think I mean, disjoint. So cases A,B are disjoint.

    P(A U B)=P(A) + P(B) - P(A n B)

    In this case [itex]m_{A\cap B}=0[/itex] because the cases:
    I - the 1st archer will hit the target
    II - the 2nd archer will hit the target

    DO NOT have something in common.

    [tex] P(A U B) = \frac{m_A+m_B-m_{A\cap B}}{n}[/tex]

    [tex] P(A U B) = \frac{m_A+m_B-0}{n}=P(A)+P(B)[/tex]

    Is this true?

    Are those cases disjoint??
    Last edited: Jun 14, 2009
  10. Jun 14, 2009 #9


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    You're saying that if one archer hits the target the other one always misses?
  11. Jun 14, 2009 #10
    Are those cases disjoint?
  12. Jun 14, 2009 #11
    Think about this intuitively. If A and B are disjoint, then event A and event B cannot occur simultaneously. So, if A is the first archer hitting the target and B is the second archer hitting the target, [tex]P(A \cap B) = 0[/tex] means that both archers cannot simultaneously hit the target. If we know that A hit the target, then it would follow that B missed the target, so the events are not independent (unless A or B always misses).

    If both archers take one shot, then we have four possible events:
    1. Both miss
    2. Archer A hits, B misses
    3. Archer A misses, B hits
    4. Both hit.

    In this case, [tex]A \cap B[/tex] is case 4, where both archers hit. If you're going to do probability by cases, it's important that you get all of them.
  13. Jun 15, 2009 #12
    Now, I understood. Thank you very much for the help.

  14. Jun 15, 2009 #13


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    You may be confusing "mutually exclusive" with "independent". "Independent" means what happens in one case does not affect what happens in the other- [itex]P(A\cap B)= P(A)P(B)[/itex] or, equivalently, P(A|B)= P(A). "Mutually exclusive" means [itex]P(A\cap B)= 0[/itex] so that P(A|B)= 0. Not at all the same thing!
  15. Jun 15, 2009 #14
    Yes, you're right. I did a little research, and find out that the events in this case aren't exclusive, but they are independent. Because if they are exclusive P(A) or P(B) will be equal 0 so that P(A n B)=0.
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