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  1. #1
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    Question on Probability

    Given that
    $\displaystyle P(\neg A)=0.6$
    $\displaystyle P(B\mid A)=0.7$
    $\displaystyle P(B)=0.3$
    What is $\displaystyle P(\neg A \wedge B)$?
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  2. #2
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    Quote Originally Posted by quiney View Post
    Given that
    $\displaystyle P(\neg A)=0.6$
    $\displaystyle P(B\mid A)=0.7$
    $\displaystyle P(B)=0.3$
    What is $\displaystyle P(\neg A \wedge B)$?
    A simple approach would be to draw a tree diagram. Have you tried doing that?
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  3. #3
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    Here is another way.
    Recall that $\displaystyle P(B) = P(B \cap A) + P(B \cap \neg A)$
    Solve for $\displaystyle P(B \cap \neg A) $.
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  4. #4
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    Hello, quiney!

    Yet another approach . . .


    Given that:
    . . $\displaystyle \begin{array}{ccc}P(\sim\!A)&=& 0.6 \\
    P(B\,|\,A) &=& 0.7 \\
    P(B) &=& 0.3 \end{array}$

    What is .$\displaystyle P(\sim\!A \wedge B)$ ?

    We can place the data into a chart:

    . . $\displaystyle \begin{array}{c||c|c||c}
    & B & \sim\!B & \text{total} \\ \hline \hline
    A & & & 0.40 \\ \hline
    \sim\!A & & & 0.60 \\ \hline \hline
    \text{Total} & 0.30 & 0.70 & 1.00 \end{array}$


    We have: .$\displaystyle P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(B \wedge A)}{P(A)} \:=\:0.7$

    . . . . $\displaystyle P(B \wedge A) \:=\:0.7\cdot P(A) \;=\;(0.7)(0.4)$

    . . . . $\displaystyle P(A \wedge B) \;=\;0.28$


    Insert that into the chart.
    . . Fill in the rest of the chart.

    . . $\displaystyle \begin{array}{c||c|c||c}
    & B & \sim\!B & \text{total} \\ \hline \hline
    A & 0.28 & 0.12 & 0.40 \\ \hline
    \sim\!A & 0.02 & 0.58 & 0.60 \\ \hline \hline
    \text{Total} & 0.30 & 0.70 & 1.00 \end{array}$


    Therefore: .$\displaystyle P(\sim\!A \wedge B) \;=\;0.02 $



    Corrected my typo . . . Thanks, Plato!
    Last edited by Soroban; Aug 17th 2010 at 03:30 PM.
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  5. #5
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    Quote Originally Posted by Soroban View Post
    We have: .$\displaystyle P(B|A) \:=\:0.7 \quad\Rightarrow\quad \dfrac{P(A \wedge B)}{P(B)} \:=\:0.7$
    Please note the typo $\displaystyle \displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}$.
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  6. #6
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    Thanks a lot everybody. So, since $\displaystyle p(B)=p(B \wedge A) + p(B \wedge \neg A)$:
    $\displaystyle 0.3=p(B \wedge A) + p(B \wedge \neg A)$

    Since $\displaystyle p(B|A)=0.7=\dfrac{p(B \wedge A)}{p(A)=0.4} \quad\Rightarrow\quad\ 0.28=p(B\wedge A) \quad\Rightarrow\ 0.3=0.28 + p(B \wedge\neg A)$
    Therefore
    $\displaystyle 0.02=p(B \wedge\neg A)=p(\neg A \wedge B)$

    Funny, the answer guide in the book says 0.2. Must be a typo.
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