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.

Answer 1

Here total number of students n(S) = 60
Let A be the event that student opted for NCC and B be the event that the student opted for NSS.
Then $n\left(A\right)=30,n\left(B\right)=32\text{and}n(A\cap B)=24\text{Thus P (A)}=\frac{n\left(A\right)}{n\left(S\right)}=\frac{30}{60}=\frac{1}{2}P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{32}{60}=\frac{8}{15}P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{24}{60}=\frac{2}{5}\text{We know that P}(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\left(i\right)P\text{(Student opted for NCC or NSS)}=\frac{1}{2+\frac{8}{15}}-\frac{2}{5}=\frac{15+16+12}{30}=\frac{19}{30}\text{(ii) P (Student has opted neither NCC or NSS)}=P(\overline{A}\cap \overline{B})=P\overline{(A\cap B)}=1-P(A\cup B)=1-\frac{19}{30}=\frac{11}{30}\text{(iii) P (Student has opted NSS but not NCC)}=P(A^{\prime }\cup B)=P\left(B\right)-P(A\cap B)=\frac{8}{15}-\frac{2}{5}=\frac{8-6}{15}=\frac{2}{15}.$