Question

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

Answer: (D)

None of these

For mutually exclusive and exhaustive events,
$0\le \frac{1+4p}{4}\le 1,0\le \frac{1-p}{4}\le 1,0\le \frac{1-2p}{2}\le 1$
$\Rightarrow 0\le 1+4p\le 4,0\le 1-p\le 4,0\le 1-2p\le 2$
$\Rightarrow -1\le 4p\le 3,-1\le -p\le 3,-1\le -2p\le 1$
$\Rightarrow -\frac{1}{4}\le p\le \frac{3}{4},-3\le p\le 1,-\frac{1}{2}\le p\le \frac{1}{2}$
Now taking intersection of above we get $p\in [-\frac{1}{4},\frac{1}{2}]$