In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.
In a class of $60$ students, $30$ opted for $NCC$ , $32$ opted for $NSS$ and $24$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, find the probability that The student has opted neither $NCC$ nor $NSS$.
Let $A$ be the event in which the selected student has opted for $NCC$ and $B$ be the event in which the selected student has opted for $NSS$.
Total number of students $=60$
Number of students who have opted for $NCC =30$
$\therefore $ $P(A)=\frac{30}{60}=\frac{1}{2}$
Number of students who have opted for $NSS =32$
$\therefore $ $P(B)=\frac{32}{60}=\frac{8}{15}$
Number of students who have opted for both $NCC$ and $NSS = 24$
$\therefore $ $P ( A$ and $B )=\frac{24}{60}=\frac{2}{5}$
$P ($ not $A$ and not $B)$
$= P(A ^{\prime}$ and $B ^{\prime})$
$= P \left( A^{\prime} \cap B ^{\prime}\right)$
$= P ( A \cup B )^{\prime}$ $[( A^{\prime} \cap B )=( A \cup B )^{\prime}$ by De Morgan's law $)]$
$=1- P ( A \cup B )$
$=1- P ( A$ or $B )$
$=1-\frac{19}{30}$
$=\frac{11}{30}$
Thus, the probability that the selected students has neither opted for $NCC$ nor $NSS$ is $\frac{11}{30}$
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