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In a certain college, $4 \%$ of men and $1 \%$ of women are taller than $1.8 \mathrm{~m}$. Also, $60 \%$ of students are women. If a student selected at random is found to be taller than $1.8 \mathrm{~m}$, then the probability that the student being a woman is
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The correct answer is:
$3 / 11$
Let $E_1, E_2$ and $A$ be the events defined as follows:
$E_1=$ Selected student is a woman
$E_2=$ Selected student is a man
$A=$ Student is taller than $1.8 \mathrm{~m}$
We have,
$$
\begin{aligned}
P\left(E_1\right) & =\frac{60}{100}, P\left(E_2\right)=\frac{40}{100} \\
\text { and } P\left(A / E_1\right) & =\frac{1}{100}, P\left(A / E_2\right)=\frac{4}{100}
\end{aligned}
$$
By Baye's rule, we have
$$
\begin{aligned}
P\left(E_1 / A\right) & =\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \\
& =\frac{\frac{60}{100} \times \frac{1}{100}}{\frac{60}{100} \times \frac{1}{100}+\frac{40}{100} \times \frac{4}{100}}=\frac{3}{11}
\end{aligned}
$$
$E_1=$ Selected student is a woman
$E_2=$ Selected student is a man
$A=$ Student is taller than $1.8 \mathrm{~m}$
We have,
$$
\begin{aligned}
P\left(E_1\right) & =\frac{60}{100}, P\left(E_2\right)=\frac{40}{100} \\
\text { and } P\left(A / E_1\right) & =\frac{1}{100}, P\left(A / E_2\right)=\frac{4}{100}
\end{aligned}
$$
By Baye's rule, we have
$$
\begin{aligned}
P\left(E_1 / A\right) & =\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \\
& =\frac{\frac{60}{100} \times \frac{1}{100}}{\frac{60}{100} \times \frac{1}{100}+\frac{40}{100} \times \frac{4}{100}}=\frac{3}{11}
\end{aligned}
$$
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