Question

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

• A. $\left[-\frac{1}{9},\frac{1}{3}\right]$
• B. $\left[-\frac{1}{3},\frac{1}{3}\right]$
• C. $\left[\frac{1}{9}\frac{,1}{3}\right]$
• D. $\left[\frac{1}{3},1\right]$

Answer 1

Answer: (C)

As the given probabilities are probabilities of $M.E$. events
$\therefore 0\le \frac{1+3p}{3}\le 1$ and $0\le \frac{1-p}{4}\le 1,0\le \frac{1-3p}{2}\le 1$
$\Rightarrow \begin{array}{ccc}-\frac{1}{3}\le p\le \frac{2}{3}& -3\le p\le 1& -\frac{1}{3}\le p\le \frac{1}{3}\\ \left(i\right)& \left(ii\right)& \left(iii\right)\end{array}$
Again, events are $M.E$.
$\therefore 0\le \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-3p}{2}\le 1$
$\Rightarrow 0\le 4\left(1+3p\right)+3\left(1-p\right)+6\left(1-3p\right)\le 12$
$\Rightarrow -13\le 12p-3p-18p\le -1$
$\Rightarrow \frac{1}{9}\le p\le \frac{13}{9}$... (iv)
Now by (i), (ii), (iii) and ( iv ), we have$\frac{1}{9}\le p\le \frac{1}{3}$