Question

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

• A. $[\frac{1}{3},\frac{1}{2}]$
• B. $[\frac{1}{4},\frac{1}{2}]$
• C. $[\frac{1}{3},\frac{2}{3}]$
• D. $[\frac{1}{3},\frac{2}{5}]$

Answer 1

The correct option is A $[\frac{1}{3},\frac{1}{2}]$

Given probabilities are $\frac{1+3p}{3},\frac{1-p}{4}$ and $\frac{1-2p}{2}$
Now,
$0\le \frac{1+3p}{3}\le 1\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3}0\le \frac{1-p}{4}\le 1\Rightarrow -3\le p\le 10\le \frac{1-2p}{2}\le 1\Rightarrow -\frac{1}{2}\le p\le \frac{1}{2}$

From the above cases, we get
$-\frac{1}{3}\le p\le \frac{1}{2}\cdots \left(1\right)$

Also given that events are mutually exclusive, so
$0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1\Rightarrow 0\le \frac{4+12p+3-3p+6-12p}{12}\le 1\Rightarrow 0\le 13-3p\le 12\Rightarrow -13\le -3p\le -1\Rightarrow \frac{1}{3}\le p\le \frac{13}{3}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we get
$p\in [\frac{1}{3},\frac{1}{2}]$