Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Since $\frac{(1+3p)}{3},\frac{(1-p)}{4}$ and $\left(\frac{1-2p}{2}\right)$ are the probabilities of the three events, we must have
$0\le \frac{1+3p}{3}\le 1,0\le \frac{1-p}{4}\le 1$ and $0\le \frac{1-2p}{2}\le 1$
$\Rightarrow -1\le 3p\le 2,-3\le p\le 1$ and $-1\le 2p\le 1$
$\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3},-1\le p\le 1$ and $-\frac{1}{2}\le p\le \frac{1}{2}$
Also as $\frac{1+3p}{3},\frac{1-p}{4}and\frac{1-2p}{2}$ are the probabilities of three mutually exclusive events
$0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1$
$\Rightarrow 0\le 4+12p+3-3p+6-12p\le 12\Rightarrow \frac{1}{3}\le p\le \frac{13}{3}$
Thus the required value of p are such that
Max. $\{-\frac{1}{3},-3,-\frac{1}{2},\frac{1}{3}\}\le p$ $\le$ min.$\{\frac{2}{3},-1,\frac{1}{2},\frac{13}{3}\}$
$\Rightarrow \frac{1}{3}\le p\le \frac{1}{2}$