Question

A person appears in examination of subjects of physics and mathematics, each consisting of two papers. The probability of failing in one paper is $\frac{1}{3},$ independent of all other papers. If the probability that, the person fails in exactly one paper, in exactly one subject is $m,$ then the value of $162m$ is

Answer 1

There are two cases possible.
Case $1:$ Person fails in exactly one paper out of four.
Probability, $p_1={}^4{C}_1\times \frac{1}{3}\times {\left(\frac{2}{3}\right)}^3=\frac{32}{81}$

Case $2:$ Person fails in exactly one paper of one subject and both papers of the other subject.
Probability, $p_2={}^2{C}_1\cdot {}^2{C}_1\cdot \frac{1}{3}\cdot \frac{2}{3}\cdot {\left(\frac{1}{3}\right)}^2=\frac{8}{81}$
$\therefore m=p_1+p_2=\frac{40}{81}$

$\underset{–}{\text{Alternate}}:$
Let $p$ be the probability of failing in one paper of one subject.
Then $p={}^2{C}_1\cdot \frac{1}{3}\cdot \frac{2}{3}=\frac{4}{9}$
Probability of failing in more than one paper of one subject, $q=\frac{5}{9}$.
This is a binomial distribution with $n=2$.
$\therefore m={}^2{C}_1\times \frac{4}{9}\times \frac{5}{9}$
$\Rightarrow m=\frac{40}{81}$
Hence, $162m=80$