Question

If $(1+3p)/3,(1-p)/4,(1-2p)/2$ are the possibilities of three mutually exclusive events. Find, if any, the limits for the values of $p$.

• A. $p\in [0,\frac{1}{2}]$
• B. $p\in [\frac{1}{3},1]$
• C. $p\in [\frac{1}{3},\frac{1}{2}]$
• D. $p\in [\frac{1}{2},1]$

Answer 1

The correct option is C $p\in [\frac{1}{3},\frac{1}{2}]$

Since $\frac{1+3P}{3},\frac{1-P}{4}$ and $\frac{1-2P}{2}$
are the probability of mutually exclusive events.

$\therefore \frac{1+3P}{3}+\frac{1-P}{4}+\frac{1-2P}{2}\le 1$
$\Rightarrow 4+12P+3-3P+6-12P\le 12$
$\Rightarrow 13-3P\le 12\Rightarrow P\ge \frac{1}{3}$ ...(1)

and $0\le \frac{1+3P}{3}\le 1,0\le \frac{1-P}{4}\le 1,0\le \frac{1-2P}{2}\le 1$
$\Rightarrow 0\le 1+3P\le 3,0\le 1-P\le 4,0\le 1-2P\le 2$
$\Rightarrow \frac{-1}{3}\le P\le \frac{2}{3},1\ge P\ge -3,\frac{1}{2}\ge P\ge \frac{-1}{2}$
$\Rightarrow \frac{1}{3}\le P\le \frac{1}{2}$