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

A box contains 10 mangoes out of which 4 arerotten. Two mangoes are ta

1.	A box contains 10 mangoes out of which 4 arerotten. Two mangoes are taken out together. If one of them is found to begood, then find the probability that the other is also good.
Answer» 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)`

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

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