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

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(C) Total mangoes $= 10$. Good mangoes $= 6$. Rotten mangoes $= 4$.Let $G_1$ be the event that the first mango is good and $G_2$ be the event that the

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$\frac{5}{18}$

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(C) Total mangoes $= 10$. Good mangoes $= 6$. Rotten mangoes $= 4$.Let $G_1$ be the event that the first mango is good and $G_2$ be the event that the second mango is good.We are given that at least one mango is good. Let $A$ be the event that at least one mango is good.Total ways to select $2$ mangoes from $10$ is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.Ways to select $2$ rotten mangoes $= ^4C_2 = 6$.Ways to select at least one good mango $= 45 - 6 = 39$.Ways to select $2$ good mangoes $= ^6C_2 = 15$.The probability that both are good given that at least one is good is $\frac{15}{39} = \frac{5}{13}$.Wait,re-evaluating: If one is found to be good,the sample space is restricted to cases where at least one is good. The number of ways to pick at least one good mango is $39$. The number of ways to pick two good mangoes is $15$. Thus,the probability is $\frac{15}{39} = \frac{5}{13}$.

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

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