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Probabilty of AnB

  1. Feb 25, 2009 #1
    Hi Guys,
    I have been given the following Question.
    Given that P(A) = 0.4 and P(AorB) = 0.9 Find P(B) if A and B are independent.

    I know That if the're independent

    P(A&B) = P(A)P(B)

    P(AorB) = P(A) + P(B) - P(A&B)

    And as the're independent

    P(A|B) = P(A)


    P(A|B) = P(A&B)/P(B)

    As P(A&B) = P(A) + P(B) - P(A&B)

    I know P(A) = 0.4, P(A|B) = 0.4 P(AorB)= 0.9 and

    P(AorB) <= P(A) + P(B)
    0.9 <= 0.4 + P(B)
    Which implies that P(B) <= 0.5

    I'm Trying to find P(B) however I'm having a lot of difficulty finding P(A&B).

    Could someone please point me in the right direction of finding P(A&B)?

    Kindest regards
  2. jcsd
  3. Feb 25, 2009 #2


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    Science Advisor

    Typo: that last "P(A&B)" should be P(AorB)

    You have already said that P(A&B)= P(A)P(B) because they are independent.
    P(AorB)= P(A)+ P(B)- P(A&B)= P(A)+ P(B)- P(A)P(B).

    You are told that P(A)= 0.4 and P(AorB)= 0.9. That equation becomes
    0.9= 0.4+ P(B)- 0.4P(B). Solve that equation.
  4. Feb 25, 2009 #3
    Thanks alot for your help I really appreciate it
    0.9= 0.4+ P(B)- 0.4P(B). Solve that equation.

    p(B) = .833 and p(A&B) = .33

    I will be remembering that one!

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