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Homework Help: Event Probability question

  1. Feb 20, 2009 #1
    Events A and B are events such that

    P(A) = 1/2
    P(B) = 1/4
    P(A or B but not both) = 1/3.

    Find P(A intersect B)

    so far what i have in mind is that , for A to happens, it need to be 1/3 x 1/2 which is 1/6
    and for B to happens, it need to be 1/4 x 1/3 which is 1/12. Then i'm stuck here, not sure how should i find the intersect point. Please help.
     
  2. jcsd
  3. Feb 20, 2009 #2

    HallsofIvy

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    I'm not sure what you mean by "for A to happens, it need to be 1/3 x 1/2". Do you mean the probability that A happens is 1/6? No, you have already said that the probability that A happens is 1/2 alone.

    Think about this simplified scenario: A contains 6 objects, 2 of which are also in B, and B contains 5 objects. |A|= 6, |B|= 5, |A intersect B|= 2. How many objects are in A union B (in "A or B")? If we just add |A|+ |B|= 11, that's two large because we are counting objects in A intersect B twice, once in A and once in B. There are |A|+ |B|- |A intersect B|= 6+ 5- 2= 9 objects in A union B. That's were we get the rule P(A union B)= P(A)+ P(B)- P(A intersect B). Now, to get the number of objects in "A or B but not both" we just subtract A intersect B again: 9- 2= 7.
    Yes, that is correct: there are 6- 2= 4 objects in "A but not B", 5- 2= 3 objects in "B but not A" and so 4+ 3= 7 objects in A or B but not Both. The number of objects in A or B but not Both is |A|+ |B|- 2|A intersect B|.

    Converting that to probability requires only dividing through by the total number of objects so: P(A or B but not both)= P(A)+ P(B)- 2P(A intersect B).

    P(A or B but not both)= 1/3= 1/2+ 1/4- 2P(A intersect B).

    Solve for P(A intersect B).
     
  4. Feb 20, 2009 #3

    Dick

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  5. Feb 21, 2009 #4
    P(A or B but not both) = P(A or B) - P(A and B)
    P(A or B) = P(A) + P(B) - P(A and B)
    P(A or B) - P(A and B) = P(A) + P(B) - 2P(A and B)

    I think you can continue from here.
     
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