Question

There are two groups of subjects, one of which consists of $6$ Science and $4$ Engineering subjects; and the other consists of $4$ Science and $6$ Engineering subjects. An unbiased die is rolled. If number $1$ or $6$ turn up, a subject is selected at random from the first group, otherwise a subject is selected from the second group. If ultimately an engineering subject is selected, then find the probability that it is selected from second group.

• A. $\frac{3}{4}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

Let $E_1$, $E_2$ and $A$ be the events defined as follows :
$E_1 =$ die turns up with number $1$ or $6$ i.e. selecting first group of subjects
$E_2 =$ die turns up with number $2$, $3$, $4$ or $5$ i.e. selecting second group of subjects
$A =$ engineering subject is selected.
Then, $P(E_1) = \frac{2}{6} = \frac{1}{3} $
and $P(E_2) = \frac{4}{6} = \frac{2}{3} $
$P(A|E_1) = P$(selecting engineering subject from first group)
$= \frac{4}{10} = \frac{2}{5} $
$P(A|E_2) = P$(selecting engineering subject from second group)
$= \frac{6}{10} = \frac{3}{5} $
We want to find $P(E_2|A)$
By Bayes' theorem, we have
$P\left(E_{2}|A\right) = \frac{P\left(E_{2}\right)P\left(A |E_{2}\right)}{P \left(E_{1}\right)P\left(A|E_{1}\right) + P\left(E_{2}\right)P\left(A|E_{2}\right)}$
$= \frac{\frac{2}{3}\times\frac{3}{5}}{\frac{1}{3}\times \frac{2}{5}+\frac{2}{3}\times \frac{3}{5}} = \frac{\frac{2}{5}}{\frac{2}{15}+\frac{2}{5}}$
$= \frac{2}{5}\times \frac{15}{8} = \frac{3}{4}$