Question

A box contains 10 mangoes out of which 4 are rotten. Two mangoes are taken together. If one of them is found to be good, the probability that the other is rotten is
a.$\dfrac{5}{13}$
b.$\dfrac{7}{13}$
c.$\dfrac{8}{15}$
d.$\dfrac{9}{13}$

Answer 1

Hint: The number of rotten mangoes is 4 and the number of good mangoes is 6. We have a total of 10 mangoes in the box. We know the formula,$\text{Probability=}\dfrac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$ . Get the probability that the first mango is rotten and the second mango is good. Similarly, get the probability that the first mango is good and the second mango is rotten. Total probability is the summation of these probabilities.

Complete step-by-step answer:
According to the question, it is given that,
The number of rotten mangoes = 4,
The number of non-rotten mangoes = 10 – 4 = 6,
The total number of mangoes = 10.
We have to find the probability that one mango is rotten out of two mangoes. So, it may be either the first mango is rotten or the second mango is rotten.
Suppose the first mango is rotten and the other mango is good.
The probability that the first mango is rotten = $\dfrac{\text{no}\text{. of}\,\text{rotten mangoes}}{\text{total number of mangoes}}=\dfrac{4}{10}$ .
As we have drawn 1 mango from the box, so there is a total of 9 mangoes left.
The probability that the second mango is good = $\dfrac{\text{no}\text{. of}\,\text{good mangoes}}{\text{total number of mangoes}}\text{=}\dfrac{6}{9}$ .
The probability that the first mango is rotten and the other mango is good = $\dfrac{4}{10}\times \dfrac{6}{9}=\dfrac{24}{90}$ …………..(1)
Suppose the first mango is good and the other mango is rotten.
The probability that the first mango is good= $\dfrac{\text{no}\text{. of}\,\text{good mangoes}}{\text{total number of mangoes}}\text{=}\dfrac{6}{\text{10}}$ .
As we have drawn 1 mango from the box, so there is a total of 9 mangoes left.
The probability that the second mango is rotten = $\dfrac{\text{no}\text{. of}\,\text{rotten mangoes}}{\text{total number of mangoes}}\text{=}\dfrac{4}{9}$ .
The probability that the first mango is good and the other mango is rotten = $\dfrac{6}{10}\times \dfrac{4}{9}=\dfrac{24}{90}$ …………………(2)
Total probability is the summation of probabilities in equation (1) and equation (2).
Total probability = $\dfrac{24}{90}+\dfrac{24}{90}=\dfrac{48}{90}=\dfrac{8}{15}$ .
Hence, the correct option is option (C).

Note: In this question, when we are calculating the probability that the first mango is rotten and the second mango is good, one can calculate it as,
The probability that the first mango is rotten = $\dfrac{\text{no}\text{. of}\,\text{rotten mangoes}}{\text{total number of mangoes}}=\dfrac{4}{10}$ .
The probability that the second mango is good = $\dfrac{\text{no}\text{. of}\,\text{good mangoes}}{\text{total number of mangoes}}\text{=}\dfrac{6}{10}$ .
This is wrong. We can’t take the total number of mangoes as 10 when we are drawing the second mango because we already have drawn one mango in the first draw. So, we will have only 9 mangoes when we are drawing the second mango.