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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 opted for $NCC$ or $NSS$.
$\frac{19}{30}$
$\frac{19}{30}$
$\frac{19}{30}$
$\frac{19}{30}$
Solution
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 $B )=\frac{24}{60}=\frac{2}{5}$
We know that $P ( A $ or $ B )= P ( A )+ P ( B )- P ( A$ and $B )$
$\therefore $ $P ( A$ or $B )=\frac{1}{2}+\frac{8}{15}-\frac{2}{5}=\frac{15+16-12}{30}=\frac{19}{30}$
Thus, the probability that the selected student has opted for $NCC$ or $NSS$ is $\frac{19}{30}$
