Question

If $\frac{1+4p}{4},\frac{1-p}{2},\frac{1-2p}{2}$ are probabilities of three mutually exclusive events, then

• A. $\frac{1}{3}\le p\le \frac{1}{2}$
• B. $\frac{1}{3}\le p\le \frac{2}{3}$
• C. $\frac{1}{6}\le p\le \frac{1}{2}$
• D. None of these

Answer 1

The correct option is D None of these

As $\frac{1+4p}{4},\frac{1-p}{2},\frac{1-2p}{2}$ are probabilities of three mutually exclusive events, we must have

$0\le \frac{1+4p}{4}\le 1,0\le \frac{1-p}{2}\le 1,0\le \frac{1-2p}{2}\le 1$

and $0\le \frac{1+4p}{4}+\frac{1-p}{2}+\frac{1-2p}{2}\le 1$

$\Rightarrow -\frac{1}{4}\le p\le \frac{3}{4},-1\le p\le 1,-\frac{1}{2}\le p\le \frac{1}{2}$ and $\frac{1}{2}\le p\le \frac{5}{2}$

$\Rightarrow max\{-\frac{1}{4},-1,-\frac{1}{2},\frac{1}{2}\}<p\le min\{\frac{3}{4},1,\frac{1}{2},\frac{5}{2}\}$

$\Rightarrow \frac{1}{2}\le p\le \frac{1}{2}\Rightarrow p=\frac{1}{2}$