Question

If $ \frac{1+4p}{p},\,\,\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}{2}\le p\le \frac{2}{3} $
• C. $ \frac{1}{6}\le p\le \frac{1}{2} $
• D. None of these

Answer 1

Answer: (D)

Since, $ \frac{1+4p}{p},\,\,\frac{1-p}{2} $ and $ \frac{1-2p}{2} $
are probabilities of three mutually exclusive events, therefore,
$ 0\le \frac{1+4p}{p}\le 1,\,\,0\le \frac{1-p}{2}\le 0,\,\,\frac{1-2p}{2}\le 1 $
and $ 0\le \frac{1+4p}{p}+\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 $ $ \left\{ \frac{-1}{4},\,\,-1,\,\,\frac{-1}{2},\,\,\frac{1}{2} \right\}\le p\le \min \left\{ \frac{3}{4},\,\,1,\,\,\frac{1}{2},\,\,\frac{5}{2} \right\} $
$ \Rightarrow $ $ \frac{1}{2}\le p\le \frac{1}{2}\Rightarrow p=\frac{1}{2} $