Question

If $\frac{(1-3p)}{2}$, $\frac{(1+4p)}{3}$, $\frac{(1+p)}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all value of p is

• A. (0, 1)
• B. $(\frac{-1}{4},\frac{1}{3})$
• C. $(0,\frac{1}{3})$
• D. $(0,\infty )$

Answer 1

The correct option is B

$(\frac{-1}{4},\frac{1}{3})$

$P\left(A\right)=\frac{(1-3p)}{2}$

$P\left(B\right)=\frac{(1+4p)}{3}$

$P\left(C\right)=\frac{(1+p)}{6}$

The events are mutually exclusive and exhaustive.

$\therefore P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)=1$

$\Rightarrow 0\le P\left(A\right)\le 1,0\le P\left(B\right)\le 1,0\le P\left(C\right)\le 1$

$\Rightarrow 0\le \frac{1-3p}{2}\le 1,0\le \frac{1+4p}{3}\le 1,0\le \frac{1+p}{6}\le 1$

$\Rightarrow \frac{-1}{3}\le p\le \frac{1}{3}$ ....... (i)

$\frac{-1}{4}\le p\le \frac{1}{2}$ ....... (ii)

and $-1\le p\le 5$ ....... (iii)

The common solution of (i), (ii) and (iii)

is $\frac{-1}{4}\le p\le \frac{1}{3}$

$\therefore$ The set of values of p are $(\frac{-1}{4},\frac{1}{3})$