A box contains 10 mangoes,out of which 4 are | vedclass

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(C) Total mangoes $= 10$. Rotten mangoes $= 4$. Good mangoes $= 6$.We select $2$ mangoes out of $10$. The total number of ways to select $2$ mangoes i

Vedclass · Accepted Answer

$\frac{5}{13}$

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(C) Total mangoes $= 10$. Rotten mangoes $= 4$. Good mangoes $= 6$.We select $2$ mangoes out of $10$. The total number of ways to select $2$ mangoes is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.Let $A$ be the event that at least one mango is good,and $B$ be the event that both mangoes are good.We want to find the conditional probability $P(B|A) = \frac{P(B \cap A)}{P(A)}$. Since $B \subset A$,$P(B \cap A) = P(B)$.Number of ways to select $2$ good mangoes (Event $B$) $= ^6C_2 = \frac{6 	imes 5}{2} = 15$.Number of ways to select at least one good mango (Event $A$) = Total ways - Ways to select $2$ rotten mangoes $= 45 - ^4C_2 = 45 - 6 = 39$.Therefore,the required probability is $P(B|A) = \frac{15}{39} = \frac{5}{13}$.

$A$ 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,what is the probability that the other is also good?

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