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Q.
If $\frac{1 +3p}{3}, \frac{1 -p}{4}$ and $\frac{1 -3p}{2}$ are probabilities of three $M.E$. and exhaustive events, then the set of values of $p$ is
Probability
Solution:
As the given probabilities are probabilities of $M.E$. events
$\therefore 0 \le \frac{1 +3p}{3} \le1$ and $0\le \frac{1 -p}{4} \le1, 0\le \frac{1 -3p}{2} \le1$
$\Rightarrow \begin{matrix}- \frac{1}{3} \le p \le\frac{2}{3}&-3 \le p \le 1&- \frac{1}{3} \le p \le \frac{1}{3}\\ \left(i\right)&\left(ii\right)&\left(iii\right)\end{matrix}$
Again, events are $M.E$.
$\therefore 0 \le \frac{1 +3p}{3} + \frac{1 -p}{4} + \frac{1 -3p}{2} \le1$
$\Rightarrow 0\le 4\left(1 +3p\right) +3\left(1 -p\right) +6\left(1 -3p\right) \le12$
$\Rightarrow -13 \le12p -3p -18p \le -1$
$\Rightarrow \frac{1}{9} \le p \le \frac{13}{9}$... (iv)
Now by (i), (ii), (iii) and ( iv ), we have$\frac{1}{9} \le p \le \frac{1}{3}$