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Math Help - equally likely events

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    equally likely events

    if A, B, C are 3 equally likely events, what is the smallest value for P(A) such that P ( A ∩ B ∩ C ) always exceeds 0.95?
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     P(A \cap B \cap C)>0.95<br />
\iff P(A^c \cup B^c \cup C^c) \leq 0.05 (use DeMorgan's Laws)

    To find the smallest value of P(A) is equivalent to finding the largest value of P(A^c).

    Now the union of these three equal sets is largest when they are all disjoint (This can be proven formally using P(A \cup B) = P(A) + P(B) - P(A \cap B).)

    Hence, the largest value of P(A^c) is \frac{0.05}{3}, and it follows that the smallest value of P(A) is (1 - \frac{0.05}{3}).
    Last edited by h2osprey; January 30th 2010 at 01:30 AM.
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    Quote Originally Posted by stressedout View Post
    if A, B, C are 3 equally likely events, what is the smallest value for P(A) such that P ( A ∩ B ∩ C ) always exceeds 0.95?
    Since the events are independent, \Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C).
    Since the events are equally likely, \Pr(A) = \Pr(B) = \Pr(C) = p, say.

    Therefore \Pr(A \cap B \cap C) = p^3. You require the value of p such that p^3 = 0.95.
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    Quote Originally Posted by mr fantastic View Post
    Since the events are independent, \Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C).
    Since the events are equally likely, \Pr(A) = \Pr(B) = \Pr(C) = p, say.

    Therefore \Pr(A \cap B \cap C) = p^3. You require the value of p such that p^3 = 0.95.
    Hmm I think I'm missing something - why are the events independent?
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    Quote Originally Posted by h2osprey View Post
    Hmm I think I'm missing something - why are the events independent?
    My mistake. You haven't missed anything.
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