Question

If $\frac{1+3P}{3},\frac{1-P}{4}$ and $\frac{1-2P}{2}$ are the probabilities of three mutually exclusive events, then $P\in$

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

Answer 1

The correct option is D $\left[\frac{1}{3},\frac{1}{2}\right]$

Given, $P\left(E_1\right)=\frac{1+3P}{3},P\left(E_2\right)=\frac{1-P}{4},P\left(E_3\right)=\frac{1-2P}{2}$

Using $0\le P\left(E_i\right)\le 1$, we get

$0\le \frac{1+3P}{3}\le 1\Rightarrow -\frac{1}{3}\le P\le \frac{2}{3}....\left(1\right)$
$0\le \frac{1-P}{4}\le 1\Rightarrow -3\le P\le \mathrm{1.....}\left(2\right)$
$0\le \frac{1-2P}{2}\le 1\Rightarrow -\frac{1}{2}\le P\le \frac{1}{2}....\left(3\right)$
Also, $E_1,E_2,E_3$ are three mutually exclusive events.

So, $P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=\frac{1+3P}{3}+\frac{1-P}{4}+\frac{1-2P}{2}=\frac{13-3P}{12}$

We know $0\le \frac{13-3P}{12}\le 1$

$\Rightarrow 0\le 13-3P\le 12$

$\Rightarrow -13\le -3P\le -1$

$\Rightarrow \frac{1}{3}\le P\le \frac{13}{3}....\left(4\right)$

Hence, the range of values of $P$ common to $\left(1\right),\left(2\right),\left(3\right),\left(4\right)$ are between $[\frac{1}{3},\frac{1}{2}]$.