Question

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. [0 , 1]
• B. $\left[\frac{1}{3} , \frac{1}{2}\right]$
• C. $\left[\frac{1}{3} , \frac{2}{3}\right]$
• D. $\left[\frac{1}{3} , \frac{13}{3}\right]$

Answer 1

Answer: (B)

$P\left(A\right) = \frac{3x+1}{3} , P\left(B\right) = \frac{1-x}{4}, $
$ P\left(C\right) = \frac{1-2x}{2}$
$ \because$ For any event $ E, 0 \le P\left(E\right) \le 1$
$ \Rightarrow 0 \le \frac{3x+1}{3} \le1, 0 \le \frac{1-x}{4} \le1$
and $ 0 \le \frac{1-2x}{2} \le 1 $
$\Rightarrow -1 \le 3x \le 2, -3 \le x \le 1$ and $ -1 \le 2x \le1 $
$ \Rightarrow - \frac{1}{3} \le x \le \frac{2}{3} \le -3 \le x \le 1 , $
and $- \frac{1}{2} \le x \le \frac{1}{2} $
Also for mutually exclusive events A, B, C,
$ P\left(A\cup B \cup C\right) = P\left(A\right) + P\left(B\right) + P\left(C\right) $
$ \Rightarrow P\left(A\cup B \cup C\right) = \frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2}$
$ \therefore 0 \le \frac{1+3x}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \le 1$
$ 0 \le 13-3x \le 12 \Rightarrow 1 \le 3x \le 13$
$ \Rightarrow \frac{1}{3} \le x \le \frac{13}{3}$
Considering all inequations, we get
$ \max\left\{ - \frac{1}{3} , -3, - \frac{1}{2}, \frac{1}{3}\right\} \le x \le \min \left\{\frac{2}{3} , 1, \frac{1}{2} , \frac{13}{3}\right\} $
$\frac{1}{3} \le x \le \frac{1}{2} \Rightarrow x \in \left[\frac{1}{3} , \frac{1}{2} \right]$