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If $\mathrm{E}$ and $\mathrm{F}$ are events such that $P(\bar{F})=0.7$ and $P(E \cap F)=0.2$, then $\mathrm{P}(E \mid F)$ is
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The correct answer is:
$2 / 3$
$\mathrm{P}(\overline{\mathrm{F}})=0.7$ and $\mathrm{PE} \mathrm{nF})=0.2$
$P\left(\frac{E}{F}\right)=\frac{P(E n F)}{P(F)}$
Here, $\mathrm{P}(\mathrm{F})=1-\mathrm{P}(\overline{\mathrm{F}})$
$$
=1-0.7=0.3
$$
$$
P\left(\frac{E}{F}\right)=\frac{0.2}{0.3}=\frac{2}{3}
$$
$P\left(\frac{E}{F}\right)=\frac{P(E n F)}{P(F)}$
Here, $\mathrm{P}(\mathrm{F})=1-\mathrm{P}(\overline{\mathrm{F}})$
$$
=1-0.7=0.3
$$
$$
P\left(\frac{E}{F}\right)=\frac{0.2}{0.3}=\frac{2}{3}
$$
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