Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
(i) $P(A \cap B)$
(ii) $P(A \cup B)$
(iii) $P$ (A|B)
(iv) $P$ (B|A)
It is given that P (A) = 0.3 and P (B) = 0.4
(i) If A and B are independent events, then
$P(A \cap B)=P(A) \cdot P(B)=0.3 \times 0.4=0.12$
(ii) It is known that, $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.3+0.4-0.12=0.58$
(iii) It is known that, $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}$
$\Rightarrow \mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{0.12}{0.4}=0.3$
(iv) It is known that, $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}$
$\Rightarrow \mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{0.12}{0.3}=0.4$