Q. If $\frac{1 + 3 \text{p}}{3} \text{,} \frac{1 - \text{p}}{4}$ and $\frac{1 - 2 \text{p}}{2}$ are probabilities of mutually exclusive events of a random experiment, then the range of $\text{p}$ is
Solution:
Since, the probability lies between $\text{0}$ and $\text{1} \text{.}$
$0 \leq \frac{1 + 3 \text{p}}{3} \leq 1 \text{,} \, 0 \leq \frac{1 - \text{p}}{4} \leq 1 \text{,} \, 0 \leq \frac{1 - 2 \text{p}}{2} \leq 1$
$\Rightarrow 0 \leq 1 + 3 \text{p} \leq 3 \text{,} \, 0 \leq 1 - \text{p} \leq 4 \text{,} \, 0 \leq 1 - 2 \text{p} \leq 2$
$\Rightarrow \, - \frac{1}{3} \leq \text{p} \leq \frac{2}{3} \text{,} - 3 \leq \text{p} \leq 1 \text{,} \, - \frac{1}{2} \leq \text{p} \leq \frac{1}{2} \, \ldots \left(\text{i}\right)$
Again, the events are mutually exclusive
$0 \leq \frac{1 + 3 \text{p}}{3} + \frac{1 - \text{p}}{4} + \frac{1 - 2p}{2} \leq 1$
$\Rightarrow 0 \leq 1 3 - 3 \text{p} \leq 1 2$
$\Rightarrow \frac{1}{3} \leq \text{p} \leq \frac{1 3}{3} \ldots \left(\text{ii}\right)$
From Eqs. (i) and (ii), we get
$\text{max} \left\{- \frac{1}{3} , - 3,\frac{- 1}{2},\frac{1}{3}\right\} \leq \text{p} \leq \text{min} \left\{\frac{2}{3},1,\frac{1}{2},\frac{1 3}{3}\right\}$
$\Rightarrow \frac{1}{3} \leq \text{p} \leq \frac{1}{2}$
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