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Math Help - Proving Probabilities

  1. #1
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    Proving Probabilities

    I am supposed to show that if P(B) is not equal to 0, then P(A|B) is greater than or equal to 0, P(B|B) = 1, and P(A1 union A2 union A3...|B) = P(A1|B) + P(A2|B) + P(A3|B) + ... for any sequence of mutually exculsive events A1, A2, A3...

    It doesn't seem like it should be so hard... in fact, it makes perfect sense that P(B|B) = 1... but I really don't know how to show it. Please help!
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  2. #2
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    Apply the definition of conditional probability:

    P(A|B) = \frac{P(A \cap B)}{P(B)}.

    So

    P(B|B) = \frac{P(B \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1

    Also we know P(B) \geq 0 \text{ and } P(A \cap B) \geq 0 so the non-negativity of the conditional probability of A given B is easy to show.

    Now consider P(A_1 \cup A_2 ... \cup A_m |B) by definition is \frac{P((A_1 \cup A_2 ... \cup A_m ) \cap B)}{P(B)} = \frac{P((A_1 \cap B) \cup (A_2 \cap B) ... \cup (A_m  \cap B)}{P(B)}
    by basic set theory,
    = \frac{P(A_1 \cap B)}{P(B)} + ... + \frac{P(A_m \cap B)}{P(B)}
    since we assume mutual exclusivity
     = P(A_1|B) +  ... + P(A_m|B)
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