Question

There are two groups of subjects one of which consists of 5 science subjects and 3 engineering subjects and the other consists of 3 science and 5 engineering subjects. An unbaised die is cast. If number 3 or number 5 turns up, a subject is selected at random from the first group, other wise the subject is selected at random from the second group. Find the probability that an engineering subject is selected ultimately.

• A. $\frac{13}{24}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{11}{24}$

Answer 1

Answer: (A)

$\frac{13}{24}$

Let $E_1$ be the event that a subject is selected from first group.
$E_2$ the event that a subject is selected from the second group.
$E$ be the event that an engineering subject is selected.
Now the probability that die shows $3$ or $5$ is
$P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}$

$P\left(E_2\right)=\frac{1}{3}=\frac{2}{3}.$
Now probability of choosing an engineering subject from first group is
$P\left(E|E_1\right)=$ $\frac{{}^3{C}_1}{{}^8{C}_1}=\frac{3}{8}$
Similarly, $P\left(E|E_2\right)=\frac{{}^5{C}_1}{{}^8{C}_1}=\frac{5}{8}$
Hence $P\left(E\right)=P\left(E_1\right)P\left(E|E_1\right)+P\left(E_2\right)P\left(E|E_2\right)$
$=\frac{1}{3}.\frac{3}{8}+\frac{2}{3}.\frac{5}{8}$
$=\frac{13}{24}$