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Thread: Probability for Three Events with unknowns

  1. #1
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    Probability for Three Events with unknowns

    If anyone can help me solve these answers would be really appreciated cause I can't seem to get it at all

    For three events A, B, and C, we know that A
 and C
 are independent, B
 and C
 are independent, A
 and B are mutually exclusive, P(A∪C)=2/3

,P(B∪C)=3/4

,P(A∪B∪C)=11/12 Find P(A), P(B) and P(C).

    Thanks alot.
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  2. #2
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    Re: Probability for Three Events with unknowns

    $P[A \cup C] = 1-P[\bar{A}\cap \bar{C}] = 1 - P[\bar{A}]P[\bar{C}] = 1 - (1-P[A])(1-P[C]) = \dfrac 2 3$

    $(1-P[A])(1-P[C]) = \dfrac 1 3$

    $P[B \cup C] = 1-P[\bar{B}\cap \bar{C}] = 1 - P[\bar{B}]P[\bar{C}] = 1 - (1-P[B])(1-P[C]) = \dfrac 3 4$

    $(1-P[B])(1-P[C]) = \dfrac 1 4$

    $P[(A \cup B \cup C] = P[(A\cup C) \cup B] = P[(AUC)] + P[B] - P[(A\cup C)\cap B] = \dfrac 2 3 + P[B] - P[(A\cap B) \cup (B \cap C)]$

    $\dfrac {11}{12} - \dfrac 2 3 = P[B] - P[\emptyset \cup (B \cap C)] = P[B] - P[B \cap C] = P[B] - P[B]P[C]$

    $\dfrac 1 4 = P[B](1-P[C])$

    so 3 equations

    $(1-P[A])(1-P[C]) = \dfrac 1 3$

    $(1-P[B])(1-P[C]) = \dfrac 1 4$

    $ P[B](1-P[C]) = \dfrac 1 4$

    grinding through it (or letting the software grind through it) we get

    $P[A] = \dfrac 1 3$

    $P[B] = \dfrac 1 2$

    $P[C] = \dfrac 1 2$
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