Question

If $\frac{1+3p}{3},\frac{1-p}{4}$ and $\frac{1-2p}{2}$ are mutually exclusive events. Then, range of $p$ is

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

Answer 1

The correct option is A $\frac{1}{3}\le p\le \frac{1}{2}$

Since, the probability lies between $0$ and $1$.
$0\le \frac{1+3p}{3}\le 1,0\le \frac{1-p}{4}\le 1,0\le \frac{1-2p}{2}\le 1$
$\Rightarrow 0\le 1+3p\le 3,0\le 1-p\le 4,0\le 1-2p\le 2$
$\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3},-3\le p\le 1,-\frac{1}{2}\le p\le \frac{1}{2}.....\left(i\right)$
Again, the events are mutually exclusive
$0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1$
$\Rightarrow 0\le 13-3p\le 12$
$\Rightarrow \frac{1}{3}\le p\le \frac{13}{3}....\left(ii\right)$
From Eqs. (i) and (ii),
$max\{-\frac{1}{3},-3,\frac{-1}{2},\frac{1}{3}\}\le p\le min\{\frac{2}{3},1,\frac{1}{2},\frac{13}{3}\}$
$\Rightarrow \frac{1}{3}\le p\le \frac{1}{2}$.