If $\frac{1+3 P}{3}, \frac{1-2 P}{2}$ are probabilities of two mutually exclusive events, then $P$ lies in…

If $\frac{1+3 P}{3}, \frac{1-2 P}{2}$ are probabilities of two mutually exclusive events, then $P$ lies in the interval.
  1. $\left[-\frac{1}{3}, \frac{1}{2}\right]$
  2. $\left(\frac{-1}{2}, \frac{1}{2}\right)$
  3. $\left[-\frac{1}{3}, \frac{2}{3}\right]$
  4. $\left(\frac{-1}{3}, \frac{2}{3}\right)$

Solution

Given, $\begin{aligned} & 0 \leq \frac{1+3 P}{3} \leq 1 \text { and } 0 \leq \frac{1-2 P}{3} \leq 1 \\ & \Rightarrow \quad 0 \leq 1+3 P \leq 3 \text { and } 0 \leq 1-2 P \leq 2 \\ & \Rightarrow \quad-1 \leq 3 P \leq 2 \text { and }-1 \leq-2 P \leq 1 \\ & \Rightarrow \quad-\frac{1}{3} \leq P \leq \frac{2}{3} \text { and }-\frac{1}{2} \leq P \leq \frac{1}{2} \\ & \therefore \quad P \in\left[-\frac{1}{3}, \frac{1}{2}\right] \end{aligned}$

Asked in: AP EAMCET 2015

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