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If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11}$,
Find (i) $P(A \cap B)$ (ii) $P(A \mid B)$. (iii) $P(B \mid A)$
Find (i) $P(A \cap B)$ (ii) $P(A \mid B)$. (iii) $P(B \mid A)$
Solution:
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Verified Answer
Given:
$$
\mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}
$$
(i) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$$
\begin{aligned}
&\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
&\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}
\end{aligned}
$$
(ii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{4}{\frac{1}{5}}}{\frac{11}{5}}=\frac{4}{2}$.
(iii) $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}$.
$$
\mathrm{P}(\mathrm{A})=\frac{6}{11}, \mathrm{P}(\mathrm{B})=\frac{5}{11}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{7}{11}
$$
(i) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$$
\begin{aligned}
&\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
&\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}
\end{aligned}
$$
(ii) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{\frac{4}{\frac{1}{5}}}{\frac{11}{5}}=\frac{4}{2}$.
(iii) $\mathrm{P}(\mathrm{B} \mid \mathrm{A})=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{\frac{4}{11}}{\frac{6}{11}}=\frac{4}{6}=\frac{2}{3}$.
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