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Math Help - Probability: Disjoint & Independence

  1. #1
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    Probability: Disjoint & Independence

    A girl is to attend 2 meetings in 24 hours. The probability she is
    tardy on the 1st meet is .4 , the probability she is tardy on the
    2nd is .5, and the probability she is tard for both meetups is .32.

    is the event she is tardy for the first independent of being tardy for the second? are they disjoint?

    what is the probability she is late for at least 1? exactly 1?

    ATTEMPT AT WORK

    P(A and B) = 0.32
    P(A)P(B) = 0.2

    THEREFORE NOT DISJOINT OR INDEPENDENT.

    EXACTLY ONE MEETING

    P(A or B) - P(A and B) = 0.38

    AT LEAST ONE

    P(A or B) = 0.9
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  2. #2
    MHF Contributor harish21's Avatar
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    Note:

     P(A \cup B) = P(A)+P(B)-P(A \cap B)

     P( A_{\mbox{only}}) = P(A) - P(A \cap B)

     P(B_{\mbox{only}}) = P(B) - P(A \cap B)
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