Question

$A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4}$, respectively. If the probability of their making a common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is

• A. $\frac{1}{12}$
• B. $\frac{1}{40}$
• C. $\frac{13}{120}$
• D. $\frac{10}{13}$

Answer 1

Answer: (D)

Let $E_1$ be the event that $A$ and $B$ obtain the same answer and $E_2$ be the event that both $A$ and $B$ obtain the correct answer.
then $E_{2} \subset E_{1}$
$P(E_1) = P$(Both $A$ and $B$ obtain same answer)
i.e. $P$(answers of both $A$ and $B$ is correct)
$+ P$(both $A$ and $B$ make the same error and obtain same incorrect answer)
$= \frac{1}{3}\times\frac{1}{4}+\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\frac{1}{20}$
$= \frac{1}{12}+\frac{2}{3}\times\frac{3}{4}\times\frac{1}{20}$
$= \frac{1}{12}+\frac{1}{40}= \frac{13}{120}$
and $P\left(E_{2}\right) = P$(both $A$ and $B$ obtain correct answer)
$= \frac{1}{3}\times\frac{1}{4} = \frac{1}{12}$
$\therefore $ Required probability $= P\left(E_{2} | E_{1}\right)$
$= \frac{P\left(E_{2} \cap E_{1}\right)}{P\left(E_{1}\right)} = \frac{P\left(E_{2}\right)}{P\left(E_{1}\right)}$
$\left[\because E_{2} \subset E_{1} \Rightarrow E_{1} \cap E_{2} = E_{2}\right]$
$= \frac{1/12}{13/120} = \frac{1}{12}\times\frac{120}{13} = \frac{10}{13}$