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In a class of 60 students, 30 opted for $N C C, 32$ opted for $N S S$ and 24 opted for both $N C C$ and $N S S$. If one of these student is selected at random. Find the probability that
(i) The student opted for $N C C$ or $N S S$.
(ii) The student has opted neither $N C C$ nor $N S S$.
(iii) The student has opted $N S S$ but not $N C C$.
(i) The student opted for $N C C$ or $N S S$.
(ii) The student has opted neither $N C C$ nor $N S S$.
(iii) The student has opted $N S S$ but not $N C C$.
Solution:
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Verified Answer
In a class of 60 students, 30 students opted for $N C C$.
$\therefore \quad$ Probability of opting $N C C=\frac{30}{60}$
Let $A$ be the event that a student opts $N C C$.
$\therefore \quad P(A)=0.5$
Let $B$ be the event that a student opts $N S S$.
$n(B)=32 \quad \therefore \quad P(B)=\frac{32}{60}$
24 students opt $N C C$ and $N S S$ both, $P(A \cap B)=\frac{24}{60}$
(i) Probability that a student opts $N S S$ or $N C C$
$\begin{aligned}
=& P(A \cup B) \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
=& \frac{30}{60}+\frac{32}{60}-\frac{24}{60}=\frac{30+32-24}{60}=\frac{38}{60}=\frac{19}{30}
\end{aligned}$
(ii) Probability that the student has opted neither $N C C$ nor $N S S$
$=P(A \cap B)^{\prime}=1-P(A \cup B)=1-\frac{19}{30}=\frac{11}{30}$
(iii) Probability that the student has opted NSS but not
$\begin{aligned}
&N C C \\
&=P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B) \\
&=\frac{32}{60}-\frac{24}{60}=\frac{8}{60}=\frac{2}{15}
\end{aligned}$
$\therefore \quad$ Probability of opting $N C C=\frac{30}{60}$
Let $A$ be the event that a student opts $N C C$.
$\therefore \quad P(A)=0.5$
Let $B$ be the event that a student opts $N S S$.
$n(B)=32 \quad \therefore \quad P(B)=\frac{32}{60}$
24 students opt $N C C$ and $N S S$ both, $P(A \cap B)=\frac{24}{60}$
(i) Probability that a student opts $N S S$ or $N C C$
$\begin{aligned}
=& P(A \cup B) \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
=& \frac{30}{60}+\frac{32}{60}-\frac{24}{60}=\frac{30+32-24}{60}=\frac{38}{60}=\frac{19}{30}
\end{aligned}$
(ii) Probability that the student has opted neither $N C C$ nor $N S S$
$=P(A \cap B)^{\prime}=1-P(A \cup B)=1-\frac{19}{30}=\frac{11}{30}$
(iii) Probability that the student has opted NSS but not
$\begin{aligned}
&N C C \\
&=P\left(A^{\prime} \cap B\right)=P(B)-P(A \cap B) \\
&=\frac{32}{60}-\frac{24}{60}=\frac{8}{60}=\frac{2}{15}
\end{aligned}$
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