Question

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

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

Answer 1

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

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