Conditional Probability

**jordanrs** · May 11th 2010, 03:25 AM

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

**Anonymous1** · May 11th 2010, 06:34 AM

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

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

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

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

**Soroban** · May 11th 2010, 10:06 AM

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 . . .

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

Fill in the empty cells:

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

Answer the questions:

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

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

**Anonymous1** · May 11th 2010, 10:14 AM

Originally Posted by Soroban

Hello, jordanrs!

Place the information in a chart . . .

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

Fill in the empty cells:

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

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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