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

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(D) Total mangoes $= 10$. Good mangoes $= 6$. Spoiled mangoes $= 4$.Let $E$ be the event that at least one mango is good, and $F$ be the event that bo

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$\frac{5}{13}$

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(D) Total mangoes $= 10$. Good mangoes $= 6$. Spoiled mangoes $= 4$.Let $E$ be the event that at least one mango is good, and $F$ be the event that both mangoes are good.The number of ways to select $2$ mangoes from $10$ is $^{10}C_2 = \frac{10 	imes 9}{2} = 45$.The number of ways to select $2$ spoiled mangoes is $^4C_2 = \frac{4 	imes 3}{2} = 6$.The number of ways to select at least one good mango is $45 - 6 = 39$. So, $P(E) = \frac{39}{45}$.The number of ways to select $2$ good mangoes is $^6C_2 = \frac{6 	imes 5}{2} = 15$. So, $P(F) = \frac{15}{45}$.Since $F \subset E$, $P(E \cap F) = P(F) = \frac{15}{45}$.The conditional probability that the other is good given that one is good is $P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{15/45}{39/45} = \frac{15}{39} = \frac{5}{13}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

$A$ box contains $10$ mangoes, out of which $4$ are spoiled. $2$ mangoes are taken together at random. If one of them is found to be good, then the probability that the other is also good is:

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