Given two independent events $A$ and $B$ such that $P(A) = 0.3,P(B) = 0.6$ . Find

(i) $P(A{\rm{ and }}B)$

(ii) $P(A{\rm{ and not }}B)$

(iii) $P(A{\rm{ or }}B)$

(iv) $P$ (neither $A$ nor $B$ )

.: (i) $P(A{\rm{ and }}B) = P(A \cap B) = P(A) \times P(B)$

( $A$ and $B$ are independent)
$\Rightarrow$ $P(A \cap B) = 0.3 \times 0.6 = 0.18$

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

( $A$ and $B$ are independent, $A$ and ${B^c}$ are also independent)

$= (0.3)(1 - P(B)) = (0.3)(1 - 0.6) = (0.3)(0.4) = 0.12$

(iii) $P(A{\rm{ or }}B) = P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\Rightarrow$ $P(A \cup B) = P(A) + P(B) - P(A)P(B)$

( $A$ and $B$ are independent)

$= 0.3 + 0.6 - 0.3 \times 0.6 = 0.9 - 0.18 = 0.72$

( $P$ (neither $A$ nor $B$ ) $= P\left( {{A^c}{\rm{ and }}{B^c}} \right) = P\left( {{A^c} \cap {B^c}} \right) = P{(A \cup B)^c}$ )
$= 1 - P(A \cup B) = 1 - 0.72 = 0.28$

(using part (iii))