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