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 values of p is
(a) (0, 1)
(b) (−1/4, 1/3)
(c) (0, 1/3)
(d) (0, ∞)

Answer 1

(b) (−1/4, 1/3)

P(A) = (1 − 3p)/2
P(B) = (1 + 4p)/3
P(C) = (1 + p)/6

The events are mutually exclusive and exhaustive.
∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

⇒ $0\le \mathrm{P}\left(\mathrm{A}\right)\le 1$, $0\le \mathrm{P}\left(\mathrm{B}\right)\le 1$, $0\le \mathrm{P}\left(\mathrm{C}\right)\le 1$

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

$\Rightarrow -1/3\le p\le \frac{1}{3}$, ...(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 $-1/4\le p\le 1/3$.
∴ The set values of p are ($-$1/4 , 1/3)