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

The correct option is A

$\frac{1}{3}\le p\le \frac{1}{2}$

Step-1: Express the given data.

$\frac{1+3p}{3},\frac{1-p}{4},\frac{1-2p}{2}$are the probability of the three events.

$\Rightarrow$$0\le \frac{1+3p}{3}\le 1,$$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},$ $-3\le p\le 1,$ $-\frac{1}{2}\le p\le \frac{1}{2}.$

Step-2: Apply condition for mutually exclusive.

$\Rightarrow$$0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1$

$\Rightarrow$$0\le \frac{4\left(1+3p\right)+3\left(1-p\right)+6\left(1-2p\right)}{12}\le 1$

$\Rightarrow$$0\le \frac{-3p+13}{12}\le 1$

$\Rightarrow$$0\le -3p+13\le 12$

$\Rightarrow$$\frac{1}{3}\le p\le \frac{13}{3}$

Therefore, minimum value $\left\{-\frac{1}{3},-3,-\frac{1}{2},\frac{1}{3}\right\}$ and maximum value $\left\{\frac{2}{3},1,\frac{1}{2},\frac{13}{3}\right\}$.

Thus, $\frac{1}{3}\le p\le \frac{1}{2}$

Hence, correct answer is option $\left(A\right)$