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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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$\frac{5}{13}$

Vedclass · Answer

(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 $P(F|E) = \frac{P(F \cap E)}{P(E)} = \frac{15/45}{39/45} = \frac{15}{39} = \frac{5}{13}$.

List-$I$	List-$II$
$A$. $P(\frac{A}{B})$	$(i)$. $\frac{2}{15}$
$B$. $P(\bar{B})$	$(ii)$. $\frac{4}{15}$
$C$. $P(A \cap \bar{B})$	$(iii)$. $\frac{8}{15}$
$D$. $P(B \cap \bar{A})$	$(iv)$. $\frac{2}{3}$
	$(v)$. $\frac{3}{7}$

$A$ 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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