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A and B are two events such that $\bar{A}$ and $\bar{B}$ are mutually exclusive. If $\mathrm{P}(\mathrm{A})=0.5$ and $\mathrm{P}(\mathrm{B})=0.6$, then what is the value of $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ ?
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The correct answer is:
$\frac{1}{6}$
$\bar{A}$ and $\bar{B}$ are mutually exclusive $\therefore P(\bar{A} \cap \bar{B})=0$
Given, $\mathrm{P}(\mathrm{A})=0.5 \Rightarrow P(\bar{A})=1-0.5=0.5$
$\mathrm{P}(\mathrm{B})=0.6 \Rightarrow P(\bar{B})=1-0.6=0.4$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1-P(\bar{A} \cup \bar{B})}{P(B)}$
$=\frac{1-(P(\bar{A})+P(\bar{B}))}{P(B)}=\frac{1-(0.5+0.4)}{0.6}$
$=\frac{1-0.9}{0.6}=\frac{0.1}{0.6}=\frac{1}{6}$
Given, $\mathrm{P}(\mathrm{A})=0.5 \Rightarrow P(\bar{A})=1-0.5=0.5$
$\mathrm{P}(\mathrm{B})=0.6 \Rightarrow P(\bar{B})=1-0.6=0.4$
$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{1-P(\bar{A} \cup \bar{B})}{P(B)}$
$=\frac{1-(P(\bar{A})+P(\bar{B}))}{P(B)}=\frac{1-(0.5+0.4)}{0.6}$
$=\frac{1-0.9}{0.6}=\frac{0.1}{0.6}=\frac{1}{6}$
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