**Question:**

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

(i) The student opted for NCC or NSS.

(ii) The student has opted neither NCC nor NSS.

(iii) The student has opted NSS but not NCC.

**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

$\therefore \mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\frac{24}{60}=\frac{2}{5}$

(i) 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 $\frac{19}{30}$.

(ii)

$P$ (not $A$ and not $B$ )

$=\mathrm{P}\left(\mathrm{A}^{\prime}\right.$ and $\left.\mathrm{B}^{\prime}\right)$

$=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$

$=\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime} \quad\left[\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B})^{\prime}(\right.$ by De Morgan's law $\left.)\right]$

$=1-P(A \cup B)$

$=1-\mathrm{P}(\mathrm{A}$ or $\mathrm{B})$

$=1-\frac{19}{30}$

$=\frac{11}{30}$

Thus, the probability that the selected students has neither opted for NCC nor NSS is $\frac{11}{30}$.

(iii) The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

= *n*(B – A) = *n*(B) – *n*(A ∩ B) = 32 – 24 = 8

Thus, the probability that the selected student has opted for NSS but not for NCC $=\frac{8}{60}=\frac{2}{15}$

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