Question

At a certain university, $4$% of men are over $6$ feet tall and $1$% of women are over 6 feet tall. The total student population is divided in the ratio $3:2$ in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?

• A. $\frac{3}{11}$
• B. $\frac{5}{11}$
• C. $\frac{7}{11}$
• D. none of these

Answer 1

Answer: (A)

$\frac{3}{11}$

Let M={Student is Male}, F={Student is Female}.

Note that M and F partition the sample space of students.

Let T={Student is over 6 feet tall}.

We know that $P\left(M\right)=\frac{2}{5},P\left(F\right)=\frac{3}{5},P\left(T/M\right)=\frac{4}{100}$ and $P\left(T/F\right)=\frac{1}{100}$.

We require $P\left(F/T\right)$.

Using Bayes’ theorem we have:

$P\left(F/T\right)=\frac{P\left(T/F\right)P\left(F\right)}{P\left(T/F\right)P\left(F\right)+P\left(T/M\right)P\left(M\right)}⟹=\frac{\frac{1}{100}\times \frac{3}{5}}{\frac{1}{100}\times \frac{3}{5}+\frac{4}{100}\times \frac{2}{5}}=\frac{3}{500}\times \frac{500}{3+8}=\frac{3}{11}$