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Math Help - Is an answer possible with the given information?

  1. #1
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    Is an answer possible with the given information?

    P(A U B) = .7 P(A U compliment of B) = .9

    Find P(A)


    I've racked my brain and cannot come up with an answer.

    This is all I've managed to calculate, perhaps I'm just missing a connection.

    P(A U compliment of B) = P(compliment of B)
    P(B) = 1 - P(compliment of B) = .1
    P(compliment of A intersect compliment of B) = 1 - P(A U B) = .3
    P(A intersect compliment of B) = P(A U B) - P(B) = .6
    P(A|compliment of B) =
    P(A intersect compliment of B) / P(compliment of B) = 2/3
    "This last one establishes dependence."

    That is all I could come up with.
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  2. #2
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    Quote Originally Posted by bartoncl View Post
    P(A U B) = .7 P(A U compliment of B) = .9

    Find P(A)


    I've racked my brain and cannot come up with an answer.

    This is all I've managed to calculate, perhaps I'm just missing a connection.

    P(A U compliment of B) = P(compliment of B)
    P(B) = 1 - P(compliment of B) = .1
    P(compliment of A intersect compliment of B) = 1 - P(A U B) = .3
    P(A intersect compliment of B) = P(A U B) - P(B) = .6
    P(A|compliment of B) =
    P(A intersect compliment of B) / P(compliment of B) = 2/3
    "This last one establishes dependence."

    That is all I could come up with.
    [Note on spelling. Compliment is when you say something flattering about someone. Complement is "that which completes or makes up the whole".]

    I'll write A^c,\ B^c for the complements of A and B.

    P(A^c\cap B^c) = 1 - P(A\cup B) = 1 - .7 = .3 (you already got that one). In the same way,
    P(A^c\cap B) = 1 - P(A\cup B^c) = 1 - .9 = .1. Then P(A^c) = P(A^c\cap B) + P(A^c\cap B^c) = .3 + .1 = .4. Finally, P(A) = 1- P(A^c) = .6.
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  3. #3
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    Hello, bartoncl!

    P(A \cup B) \:=\: 0.7 \qquad P(A \cup B') \:=\: 0.9

    Find P(A).
    Draw a two-circle Venn diagram . . . it will help.


    We are given: . P(A \cup B') \:=\:0.9

    Its complement is: . P(A \cup B')' \:=\:1 - 0.9 \quad\Rightarrow\quad P(A' \cap B) \:=\:0.1


    \text{Formula: }\;P(A \cup B) \;=\;P(A) + \underbrace{P(B) - P(A \cap B)}_{\text{This is }P(A'\,\cap\,B)}

    \text{So we have: }\;\underbrace{P(A \cup B)}_{0.7} \;=\;P(A) + \underbrace{P(A' \cap B)}_{0.1}


    Therefore: . 0.7 \:=\:P(A) + 0.1 \quad\Rightarrow\quad \boxed{P(A) \:=\:0.6}

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