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Thread: equally likely events

  1. #1
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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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    $\displaystyle P(A \cap B \cap C)>0.95
    \iff P(A^c \cup B^c \cup C^c) \leq 0.05 $ (use DeMorgan's Laws)

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

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

    Hence, the largest value of $\displaystyle P(A^c)$ is $\displaystyle \frac{0.05}{3}$, and it follows that the smallest value of $\displaystyle P(A)$ is $\displaystyle (1 - \frac{0.05}{3})$.
    Last edited by h2osprey; Jan 30th 2010 at 12:30 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, $\displaystyle \Pr(A \cap B \cap C) = \Pr(A) \cdot \Pr(B) \cdot \Pr(C)$.
    Since the events are equally likely, $\displaystyle \Pr(A) = \Pr(B) = \Pr(C) = p$, say.

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

    Therefore $\displaystyle \Pr(A \cap B \cap C) = p^3$. You require the value of p such that $\displaystyle p^3 = 0.95$.
    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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