Question

In a class of $60$ students, $30$ opted for Mathematics, $32$ opted for Biology and $24$ opted for both Mathematics and Biology. If one of these students is selected at random, find the probability that:
(i) The student opted for Mathematics or Biology.
(ii) The student has opted neither Mathematics nor Biology.
(iii) The student has opted Mathematics but not Biology.

Answer 1

Number of students opted mathematics $=n\left(M\right)=30$

Number of students opted Biology $=n\left(B\right)=32$

Number of students opted both Mathematics and Biology $=n(M\cap B)=24$

Number of students opted Mathematics or Biology $=(M\cup B)$

$n(M\cup B)=n\left(M\right)+n\left(B\right)-n(M\cap B)\Rightarrow n(M\cup B)=30+32-24=38$

(i) Probability that student opted Mathematics or Biology $=\frac{n(M\cup B)}{n}=\frac{38}{60}=\frac{19}{30}$

(ii) Probability that student has opted neither mathematics nor Biology $=1-P(M\cup B)=1-\frac{19}{30}=\frac{11}{30}$

Number os students opted only Mathematics $=n\left(M\right)-n(M\cap B)=30-24=8$

(iii) Probability of selecting students who opted only mathematics $=\frac{8}{60}=\frac{2}{15}$