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If $P(A)=0 \cdot 8, P(B)=0 \cdot 5$ and $P(B / A)=0.4$, find
(i) $P(A \cap B)$
(ii) $\mathrm{P}(\mathrm{A} / \mathrm{B})$
(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
(i) $P(A \cap B)$
(ii) $\mathrm{P}(\mathrm{A} / \mathrm{B})$
(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
Solution:
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Verified Answer
(i) $\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} \Rightarrow 0 \cdot 4=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{0 \cdot 8}$
$\therefore \quad P(A \cap B)=0.4 \times 0.8=0.32$
(ii) $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}$
(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$$
=0.8+0.5-0.32=1.30-0.32=0.98
$$
$\therefore \quad P(A \cap B)=0.4 \times 0.8=0.32$
(ii) $\mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.32}{0.5}=\frac{32}{50}=\frac{16}{25}$
(iii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$$
=0.8+0.5-0.32=1.30-0.32=0.98
$$
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