In a class of 60 students,40 opted for NCC,30 | vedclass

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Question

(A) Let $A$ be the set of students who opted for $NCC$ and $B$ be the set of students who opted for $NSS$.Given: $n(U) = 60$,$n(A) = 40$,$n(B) = 30$,$

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$\frac{1}{6}$

Vedclass · Answer

(A) Let $A$ be the set of students who opted for $NCC$ and $B$ be the set of students who opted for $NSS$.Given: $n(U) = 60$,$n(A) = 40$,$n(B) = 30$,$n(A \cap B) = 20$.The number of students who opted for at least one of $NCC$ or $NSS$ is given by $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.$n(A \cup B) = 40 + 30 - 20 = 50$.The number of students who opted for neither $NCC$ nor $NSS$ is $n(A^c \cap B^c) = n(U) - n(A \cup B) = 60 - 50 = 10$.The probability that the selected student has opted for neither is $\frac{n(A^c \cap B^c)}{n(U)} = \frac{10}{60} = \frac{1}{6}$.

In a class of $60$ students,$40$ opted for $NCC$,$30$ opted for $NSS$ and $20$ opted for both $NCC$ and $NSS$. If one of these students is selected at random,then the probability that the student selected has opted neither for $NCC$ nor for $NSS$ is

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