Question

If $\frac{1-3p}{2},\frac{(1+4p)}{3},\frac{1+p}{6}$ are the probabilities of three mutually exclusive and exhaustive events,then the set of all values of p is

• A. $(0,1)$
• B. $[-\frac{1}{4},\frac{1}{3}]$
• C. $[0,\frac{1}{3}]$
• D. $(0,\infty )$

Answer 1

The correct option is B $[-\frac{1}{4},\frac{1}{3}]$

For mutually exclusive and exhaustive events,
$0\le \frac{1-3p}{2}\le 1,0\le \frac{1+4p}{3}\le 1,0\le \frac{1+p}{6}\le 1$ and $0\le \frac{1-3p}{2}+\frac{1+4p}{3}+\frac{1+p}{6}\le 1$
$\Rightarrow 0\le 1-3p\le 2,0\le 1+4p\le 3,0\le 1+p\le 6$ and $0\le \frac{3(1-3p)+2(1+4p)+1+p}{6}\le 1$
$\Rightarrow -1\le -3p\le 1,-1\le 4p\le 2,-1\le p\le 5$ and $0\le 1\le 1$
$\Rightarrow -\frac{1}{3}\le p\le \frac{1}{3},-\frac{1}{4}\le p\le \frac{1}{2},-1\le p\le 5$
Now taking intersection of above we get $p\in [-\frac{1}{4},\frac{1}{3}]$