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Let $A$ and $B$ independent events $\mathrm{P}(\mathrm{A})=0.3$ and $P(B)=0.4$. Find
(i) $\mathbf{P}(\mathbf{A} \cap \mathrm{B})$
(ii) $\mathbf{P}(\mathbf{A} \cup \mathrm{B})$
(iii) $P(A \mid B)$
(iv) $\mathbf{P}(\mathbf{B} \mid \mathbf{A})$
(i) $\mathbf{P}(\mathbf{A} \cap \mathrm{B})$
(ii) $\mathbf{P}(\mathbf{A} \cup \mathrm{B})$
(iii) $P(A \mid B)$
(iv) $\mathbf{P}(\mathbf{B} \mid \mathbf{A})$
Solution:
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Verified Answer
$P(A)=0.3, P(B)=0.4$
$A$ and $B$ are independent events.
(i) $\quad \therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})=0.3 \times 0.4=0.12$.
(ii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B})$ $=0.3+0.4-0.3 \times 0.4=0.7-0.12=0.58$.
(iii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{P(A \cap B)}{P(B)}=\frac{P(A) \cdot P(B)}{P(B)}=0.3$
(iv) $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{P(A \cap B)}{P(A)}=\frac{P(A) \cdot P(B)}{P(A)}=0.4$
$A$ and $B$ are independent events.
(i) $\quad \therefore \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})=0.3 \times 0.4=0.12$.
(ii) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) . \mathrm{P}(\mathrm{B})$ $=0.3+0.4-0.3 \times 0.4=0.7-0.12=0.58$.
(iii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{P(A \cap B)}{P(B)}=\frac{P(A) \cdot P(B)}{P(B)}=0.3$
(iv) $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{P(A \cap B)}{P(A)}=\frac{P(A) \cdot P(B)}{P(A)}=0.4$
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