Given two independent events A and B such that

Solution:

It is given that P (A) = 0.3 and P (B) = 0.6

Also, A and B are independent events.

(i) $\therefore \mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

$\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3 \times 0.6=0.18$

(ii) $P(A$ and not $B)=P\left(A \cap B^{\prime}\right)$

$=P(A)-P(A \cap B)$

$=0.3-0.18$

$=0.12$

(iii) $P(A$ or $B)=P(A \cup B)$

$=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$

$=0.3+0.6-0.18$

$=0.72$

(iv) $P$ (neither $A$ nor $B$ ) $=P\left(A^{\prime} \cap B^{\prime}\right)$

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

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

$=1-0.72$

$=0.28$