Question

If $\frac{(1+4p)}{4},\frac{(1-p)}{2},\text{and}\frac{(1-2p)}{2}$ are the probabilities of three mutually exclusive events , then the value of p is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

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

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


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

$0\le \frac{1-2p}{2}\le 1and0\le \left(\frac{1+4p}{4}\right)+\left(\frac{1-p}{2}\right)+\frac{1-2p}{2}\le 1$

$\therefore \frac{-1}{4}\le p\le \frac{3}{4}1-\le p\le 1,\frac{1}{2}\le p\le \frac{1}{2}andp\le \frac{5}{2}$


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