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.
Text Solution
Verified by Experts
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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