Q. Events $A,B,C$ are mutually exclusive events such that $P(A) = \frac{3x + 1}{3},P(B) = \frac{1 - x}{4}$and $P(C) = \frac{1 - 2x}{2}$. The set of possible values of $x$ are in the interval
A) $\left\lbrack \frac{1}{3},\frac{1}{2} \right\rbrack$
B) $\left\lbrack \frac{1}{3},\frac{2}{3} \right\rbrack$
C) $\left\lbrack \frac{1}{3},\frac{13}{3} \right\rbrack$
D) [0, 1]
Explanation
(a)
Since, $0 \leq P(A) \leq 1,\ 0 \leq P(B) \leq 1,\ 0 \leq P(C) \leq 1$
and $0 \leq P(A) + P(B) + P(C) \leq 1$
$\therefore\ 0 \leq \frac{3x + 1}{3} \leq 1$
$\Rightarrow \ - \frac{1}{3} \leq x \leq \frac{2}{3}$ ...(i)
$0 \leq \frac{1 - x}{4} \leq 1\ \Rightarrow \ - 3 \leq x \leq 1$ ...(ii)
$0 \leq \frac{1 - 2x}{2} \leq 1\ \Rightarrow \ - \frac{1}{2} \leq x \leq \frac{1}{2}$ ...(iii)
and $0 \leq \frac{3x + 1}{3} + \frac{1 - x}{4} + \frac{1 - 2x}{2} \leq 1$
$\Rightarrow \ \frac{1}{3} \leq x \leq \frac{13}{3}$ ...(iv)
From Eqs. (i), (ii), (iii) and (iv), we get
$\frac{1}{3} \leq x \leq \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.