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 $=P\left(E_2|E_1\right)$
$=\frac{P\left(E_2\cap E_1\right)}{P\left(E_1\right)}=\frac{P\left(E_2\right)}{P\left(E_1\right)}$
$\left[\because E_2\subset E_1\Rightarrow E_1\cap E_2=E_2\right]$
$=\frac{1/12}{13/120}=\frac{1}{12}\times \frac{120}{13}=\frac{10}{13}$