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

Explanation

Let A be the event that the first mango is good, and B be the event that the second one is good. Then, required probability is
`P(B//A)=(P(AnnB))/(P(A))`
Now, probability that both mangoes are goos is
`P(AnnB)=(""^(6)C_(2))/(""^(10)C_(2))`
Probability that first mango is good is
`P(A)=(""^(6)C_(2))/(""^(10)C_(2))+(""^(6)C_(1)xx^(4)C_(1))/(""^(10)C_(2))`
Hence, `P(B//A)=(""^(6)C_(2))/(""^(6)C_(2)+""^(6)C_(1)xx^(4)C_(1))=(15)/(15+24)=(5)/(13)`