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
$P\left(\frac{A}{B}\right)=\frac{\frac{{}^6{C}_2}{{}^{10}{C}_2}}{\frac{{}^6{C}_2}{{}^{10}{C}_2}+\frac{{}^6{C}_1{\times }^4{C}_1}{{}^{10}{C}_2}}$
$\Rightarrow P\left(\frac{B}{A}\right)=\frac{{}^6{C}_2}{{}^6{C}_2{+}^6{C}_1{\times }^4{C}_1}$
$\Rightarrow P\left(\frac{B}{A}\right)=\frac{15}{15+24}=\frac{15}{39}$
$\left[{\because }^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right]$
$\Rightarrow P\left(\frac{B}{A}\right)=\frac{5}{13}$