Question

A box contains 10 mangoes out of which 4 are rotten. 2 mangoes are taken out together. If one of them is found to be good, the probability that the other is also good is :

• A. $\frac{2}{3}$
• B. $\frac{5}{13}$
• C. $\frac{8}{13}$
• D. $\frac{7}{13}$

Answer 1

Answer: (B)

Let A be event that the first mango is good and B denotes the event that the second is good.
Then, required probability
$= P\left(\frac{B}{A} \right) = \frac{P\left(A \cap B\right)}{P\left(A\right)}$ ... (i)
Now, $P\left(A \cap B\right) = $ Probability that both mangoes are good.
$\Rightarrow P\left(A \cap B\right) = \frac{^{6}C_{2}}{^{10}C_{2}} $
$P\left(A\right)$ = prob. that first mango is good
$ = \frac{^{6}C_{2}}{^{10}C_{2}} + \frac{^{6}C_{1} \times ^{4}C_{1}}{^{10}C_{2}} $
$\therefore $ equation (1) becomes
$P\left(\frac{A}{B}\right) = \frac{\frac{^{6}C_{2}}{^{10}C_{2}}}{\frac{^{6}C_{2}}{^{10}C_{2}} + \frac{^{6}C_{1} \times^{4}C_{1}}{^{10}C_{2}}} $
$\Rightarrow P\left(\frac{B}{A}\right) = \frac{^{6}C_{2}}{^{6}C_{2} + ^{6}C_{1} \times^{4}C_{1}} $
$\Rightarrow P\left(\frac{B}{A}\right) = \frac{15}{15+24} = \frac{15}{39}$
$ \left[\because ^{n}C_{r} = \frac{n!}{r!\left(n-r\right)!}\right]$
$ \Rightarrow P\left(\frac{B}{A}\right) = \frac{5}{13} $