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Math Help - Find the indicated probability?

  1. #1
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    Find the indicated probability?

    P(B)=0.5, P(A or B)=0.3, and P(A and B)=0.6, find P(A)

    How would you do this?

    The answer is {P(A)= 0.4} but how in the world do you get that? How do you do it?


    Thanks!
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  2. #2
    Moo
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    Hello !

    Do you know this formula ... ?

    \mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)

    Just draw a diagram
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  3. #3
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    What kind of diagram? Like a Venn Diagram? Where would I put all the information though?

    I don't get it
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    Quote Originally Posted by hellotarina View Post
    P(B)=0.5, P(A or B)=0.3, and P(A and B)=0.6, find P(A)
    The problem contains inconsistent data.
    For all events X~\&~Y this is true: X \subseteq Y\; \Rightarrow \;P(X) \leqslant P(Y).

    But look at the given: B \subseteq A \cup B\;\text{but} \;P(B) > P(A \cup B).
    So is impossible.
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  5. #5
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    For some reason my professor tells me the answer is P(A) = 0.4 and it is for a test which I have tomorrow and I didn't have a chance to ask about it.

    I have made sure I've typed it correctly to no avail.
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  6. #6
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    Quote Originally Posted by hellotarina View Post
    For some reason my professor tells me the answer is P(A) = 0.4 and it is for a test which I have tomorrow and I didn't have a chance to ask about it.
    Perhaps you copied it down incorrectly.
    If it were P(B)=0.5,~{\color{red}P(A\cup B)=0.6,~\&~ P(A\cap B)=.3}\text{ then }P(A)=0.4 is correct.
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  7. #7
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    Well thank you guys anyway.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Perhaps you copied it down incorrectly.
    If it were P(B)=0.5,~{\color{red}P(A\cup B)=0.6,~\&~ P(A\cap B)=.3}\text{ then }P(A)=0.4 is correct.
    I suspect the question has been copied correctly but the professor who set the question has not thought past the following equation:

    \Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)

     \Rightarrow 0.3 = 0.5 + \Pr(B) - 0.6

    \Rightarrow \Pr(B) = 0.4.

    @OP: Ask your professor how comfortable s/he is that it follows from the given data that \Pr(B \cap A') = {\color{red}-} 0.1.
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