Question

If $\frac{1 + 3 p}{3},\frac{1 - p}{4}$ and $\frac{1 - 2 p}{2}$ are probabilities of mutually exclusive events of a random experiment, then the range of $p$ is

• A. $\frac{1}{3}\leq p\leq \frac{1}{2}$
• B. $\frac{1}{4}\leq p\leq \frac{1}{2}$
• C. $\frac{1}{3}\leq p\leq \frac{2}{3}$
• D. $\frac{1}{3}\leq p\leq \frac{2}{5}$

Answer 1

Answer: (A)

Since, the probability lies between $0$ and $1.$
$0\leq \frac{1 + 3 p}{3}\leq 1, \, 0\leq \frac{1 - p}{4}\leq 1, \, 0\leq \frac{1 - 2 p}{2}\leq 1$
$\Rightarrow 0≤1+3p≤3, 0≤1−p≤4, 0≤1−2p≤2$
$\Rightarrow -\frac{1}{3}\leq p\leq \frac{2}{3},-3\leq p\leq 1,-\frac{1}{2}\leq p\leq \frac{1}{2}\ldots \left(i\right)$
Again, the events are mutually exclusive
$0\leq \frac{1 + 3 p}{3}+\frac{1 - p}{4}+\frac{1 - 2 p}{2}\leq 1$
$\Rightarrow 0\leq 13-3p≤12$
$=>\frac{1}{3}\leq p\leq \frac{13}{3}\ldots \left(ii\right)$
From equations $\left(i\right)\&\left(ii\right)$ , we get
$\max\left\{- \frac{1}{3} , - 3,\frac{- 1}{2},\frac{1}{3}\right\}\leq p\leq \min\left\{\frac{2}{3},1,\frac{1}{2},\frac{13}{3}\right\}$
$\Rightarrow \frac{1}{3}\leq p\leq \frac{1}{2}$