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Math Help - Conditional Probability

  1. #1
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    Conditional Probability

    Hello,

    just doing some conditional probability revision and im stuck on the following problems as im just confused about what values to take.

    Given that P(A) = 0.4, P(B) = 0.6 and P(B|A) = 0.8

    Find

    P(B|\overline {A}) and P(\overline {A}|\overline {B})

    Answers from the previous steps are as follows:
    P(A \cap B) = 0.32
    P(A \cup B) = 0.68
    P(A|B) = 0.533

    Cheers
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  2. #2
    Super Member Anonymous1's Avatar
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    P(B) = P(B|A)P(A) + P(B|\bar A)P(\bar A)


    \implies P(B|\bar A) = \frac{P(B) - P(B|A)P(A)}{P(\bar A)}


    P(\bar A) = P(\bar A|B)P(B) + P(\bar A|\bar B)P(\bar B)


    \implies P(\bar A|\bar B) = \frac{P(\bar A) - P(\bar A|B)P(B)}{P(\bar B)}
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  3. #3
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    Hello, jordanrs!

    \text{Given: }\;P(A) \,=\,0.4,\;\;P(B) \,=\,0.6,\;\;P(B|A) \,=\, 0.8

    \text{Find: }\;\;(a)\;P\left(B\,|\,\overline {A}\right) \qquad (b)\;P\left(\overline {A}\,|\,\overline {B}\right)

    Place the information in a chart . . .

    . . \begin{array}{c||c|c||c|}<br />
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline<br /> <br />
P(A) & 0.32  & & 0.40 \\ \hline \\[-4mm]<br />
P\left(\overline{A}\right) & & & \\ \hline \hline<br />
\text{Total} & 0.60 & & 1.00  \\ \hline<br />
\end{array}



    Fill in the empty cells:

    . . \begin{array}{c||c|c||c|}<br />
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline<br /> <br />
P(A) & 0.32  & 0.08 & 0.40 \\ \hline \\[-4mm]<br />
P\left(\overline{A}\right) & 0.28 & 0.32 & 0.60 \\ \hline \hline<br />
\text{Total} & 0.60 & 0.40 & 1.00 \\ \hline<br />
\end{array}



    Answer the questions:

    (a)\;\;P(B\,|\,\overline{A}) \;=\;\frac{P(B \wedge \overline{A})}{P(\overline{A})} \;=\; \frac{0.28}{0.40} \;=\;0.70

    (b)\;\;P(\overline{A}\,|\,\overline{B}) \;=\;\frac{P(\overline{A} \wedge \overline{B})}{P(\overline{B})} \;=\; \frac{0.32}{0.40} \;=\;0.80

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  4. #4
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, jordanrs!


    Place the information in a chart . . .

    . . \begin{array}{c||c|c||c|}<br />
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline<br /> <br />
P(A) & 0.32  & & 0.40 \\ \hline \\[-4mm]<br />
P\left(\overline{A}\right) & & & \\ \hline \hline<br />
\text{Total} & 0.60 & & 1.00  \\ \hline<br />
\end{array}



    Fill in the empty cells:

    . . \begin{array}{c||c|c||c|}<br />
& P(B) & P\left(\overline{B}\right) & \text{Total} \\ \hline \hline<br /> <br />
P(A) & 0.32  & 0.08 & 0.40 \\ \hline \\[-4mm]<br />
P\left(\overline{A}\right) & 0.28 & 0.32 & 0.60 \\ \hline \hline<br />
\text{Total} & 0.60 & 0.40 & 1.00 \\ \hline<br />
\end{array}



    Answer the questions:

    (a)\;\;P(B\,|\,\overline{A}) \;=\;\frac{P(B \wedge \overline{A})}{P(\overline{A})} \;=\; \frac{0.28}{0.40} \;=\;0.70

    (b)\;\;P(\overline{A}\,|\,\overline{B}) \;=\;\frac{P(\overline{A} \wedge \overline{B})}{P(\overline{B})} \;=\; \frac{0.32}{0.40} \;=\;0.80

    This is a really cool approach.
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