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If two events $A$ and $B$. If odds against $A$ are as $2: 1$ and those in favour of $A \cup B$ are as $3: 1$, then
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Verified Answer
The correct answer is:
$\frac{5}{12} \leq P(B) \leq \frac{3}{4}$
$\begin{array}{l}
P(A)=\frac{1}{3}, P(A \cup B)=\frac{3}{4} \\
\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\quad \leq P(A)+P(B) \\
\Rightarrow \frac{3}{4} \leq \frac{1}{3}+P(B) \\
\Rightarrow P(B) \geq \frac{5}{12}
\end{array}$
Also, $\mathrm{B} \leq \mathrm{A} \cup \mathrm{B}$
$\Rightarrow \mathrm{P}(\mathrm{B}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}$
$\therefore \frac{5}{12} \leq \mathrm{P}(\mathrm{B}) \leq \frac{3}{4}$
P(A)=\frac{1}{3}, P(A \cup B)=\frac{3}{4} \\
\therefore P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\quad \leq P(A)+P(B) \\
\Rightarrow \frac{3}{4} \leq \frac{1}{3}+P(B) \\
\Rightarrow P(B) \geq \frac{5}{12}
\end{array}$
Also, $\mathrm{B} \leq \mathrm{A} \cup \mathrm{B}$
$\Rightarrow \mathrm{P}(\mathrm{B}) \leq \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}$
$\therefore \frac{5}{12} \leq \mathrm{P}(\mathrm{B}) \leq \frac{3}{4}$
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