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If $\frac{1+3 P}{3}, \frac{1-2 P}{2}$ are probabilities of two mutually exclusive events, then $P$ lies in the interval.
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Verified Answer
The correct answer is:
$\left[-\frac{1}{3}, \frac{1}{2}\right]$
Given,
$\begin{aligned}
& 0 \leq \frac{1+3 P}{3} \leq 1 \text { and } 0 \leq \frac{1-2 P}{3} \leq 1 \\
& \Rightarrow \quad 0 \leq 1+3 P \leq 3 \text { and } 0 \leq 1-2 P \leq 2 \\
& \Rightarrow \quad-1 \leq 3 P \leq 2 \text { and }-1 \leq-2 P \leq 1 \\
& \Rightarrow \quad-\frac{1}{3} \leq P \leq \frac{2}{3} \text { and }-\frac{1}{2} \leq P \leq \frac{1}{2} \\
& \therefore \quad P \in\left[-\frac{1}{3}, \frac{1}{2}\right]
\end{aligned}$
$\begin{aligned}
& 0 \leq \frac{1+3 P}{3} \leq 1 \text { and } 0 \leq \frac{1-2 P}{3} \leq 1 \\
& \Rightarrow \quad 0 \leq 1+3 P \leq 3 \text { and } 0 \leq 1-2 P \leq 2 \\
& \Rightarrow \quad-1 \leq 3 P \leq 2 \text { and }-1 \leq-2 P \leq 1 \\
& \Rightarrow \quad-\frac{1}{3} \leq P \leq \frac{2}{3} \text { and }-\frac{1}{2} \leq P \leq \frac{1}{2} \\
& \therefore \quad P \in\left[-\frac{1}{3}, \frac{1}{2}\right]
\end{aligned}$
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