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Let $A$ and $B$ be two events such that $P(A \cap B)=\frac{1}{6}, P(A \cup B)=\frac{31}{45}$ and $P(\bar{B})=\frac{7}{10}$ then
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The correct answer is:
$A$ and $B$ are independent
Given, $P(A \cap B)=\frac{1}{6}$
$P(A \cup B)=\frac{31}{45}$
and $\quad P(\bar{B})=\frac{7}{10}$
$\therefore \quad \quad P(B)=1-\frac{7}{10}=\frac{3}{10}$
Now, $\quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow \quad \frac{31}{45}=P(A)+\frac{3}{10}-\frac{1}{6}$
$\therefore \quad P(A)=\frac{31}{45}+\frac{1}{6}-\frac{3}{10}$ $=\frac{5}{9}$
Then. $\quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}=\frac{1}{5}=\frac{3}{10}>\frac{1}{6}$
and $\quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{3}{3}}=\frac{5}{9}>\frac{1}{6}$
and, $P(A) \times P(B)=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}=P(A \cap B)$
Hence, $A$ and $B$ are independent.
$P(A \cup B)=\frac{31}{45}$
and $\quad P(\bar{B})=\frac{7}{10}$
$\therefore \quad \quad P(B)=1-\frac{7}{10}=\frac{3}{10}$
Now, $\quad P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow \quad \frac{31}{45}=P(A)+\frac{3}{10}-\frac{1}{6}$
$\therefore \quad P(A)=\frac{31}{45}+\frac{1}{6}-\frac{3}{10}$ $=\frac{5}{9}$
Then. $\quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}=\frac{1}{5}=\frac{3}{10}>\frac{1}{6}$
and $\quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{6}}{\frac{3}{3}}=\frac{5}{9}>\frac{1}{6}$
and, $P(A) \times P(B)=\frac{5}{9} \times \frac{3}{10}=\frac{1}{6}=P(A \cap B)$
Hence, $A$ and $B$ are independent.
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