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Math Help - independent

  1. #1
    Senior Member sfspitfire23's Avatar
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    independent

    P(A or B) = .60 and P(A) =.20


    How do I find P(B) given that A and B are independent?

    Since I see that we have an "or" and not an "and," I would add up the probabilities so P(B)=.40? This is also the probability if A and B were mutually exclusive too?
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  2. #2
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    Hi,

    This is my first post so you may want to check it with someone who knows something about this

    Anyway I googled the problem and it came up with this

    P(A or B) = .60 and P(A) =.20 ... How do I find P(B) given that A and B are independent? - Yahoo! Answers

    The trouble is that it gives more than one answer .5 and .4 are both suggested. Personally I find .5 more convincing. Anyway below is the best argument I found on the link.

    There's a 40% chance of not A or B
    There's an 80% chance of not A
    Thus there's a 40% chance of B but not A

    Now, given not A, there's a 50% chance of B (40%/80%)
    Since A and B are independent, there must be a 50% chance of B regardless of the condition of A.
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  3. #3
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    Hello, sfspitfire23!

    P(A \cup B) \,=\, 0.60,\;\;P(A) \,=\,0.20

    Given that A and B are independent, find P(B).

    We have this formula: . P(A \cup B) \:=\: P(A) + P(B) - P(A \cap B)

    Since A and B are independent: . P(A \cap B) \:=\:P(A)\!\cdot\!P(B)

    The formula becomes: . P(A \cup B) \;=\;P(A) + P(B) - P(A)\!\cdot\!P(B)


    Substitute: . P(A \cup B) = 0.6,\;\;P(A) = 0.2

    . . 0.6 \;=\;0.2 + P(B) - 0.2\!\cdot\!P(B)

    . . 0.4 \;=\;0.8\!\cdot\!P(B)


    Therefore: . P(B) \;=\;\frac{0.4}{0.8} \;=\;0.5

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