If and are mutually exclusive events. Then, range of P is

Engineering

Mathematics

Conditional Probability

Question

If $\frac{1 + 3 p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2 p}{2}$ are mutually exclusive events. Then, range of P is

$\frac{1}{3} \leq p \leq \frac{1}{2}$

$\frac{1}{2} \leq p \leq \frac{1}{2}$

$\frac{1}{3} \leq p \leq \frac{2}{3}$

$\frac{1}{3} \leq p \leq \frac{2}{5}$

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Since, the probability lies between 0 and 1.
$0 \leq \frac{1 + 3 p}{3} \leq 1, 0 \leq \frac{1 - p}{4} \leq 1, 0 \leq \frac{1 - 2 p}{2} \leq 1$
$\Rightarrow 0 \leq 1 + 3P \leq 3,0 \leq 1 - P \leq 4,0 \leq 1 - 2P \leq 2$
$- \frac{1}{3} \leq p \leq \frac{2}{3}, - 3 \leq p \leq 1, - \frac{1}{2} \leq p \leq \frac{1}{2} .. .. . (i)$
Again, the events are mutually exclusive
$0 \leq \frac{1 + 3 p}{3} + \frac{1 - p}{4} + \frac{1 - 2 p}{2} \leq 1$
$\Rightarrow 0 \leq 13 - 3P \leq 12$
$\frac{1}{3} \leq p \leq \frac{13}{3} .. .. (ii)$
From Eqs. (i) and (ii),
$\max {- \frac{1}{3}, - 3, \frac{- 1}{2}, \frac{1}{3}} \leq p \leq \min {\frac{2}{3}, 1, \frac{1}{2}, \frac{13}{3}}$
$\frac{1}{3} \leq p \leq \frac{1}{2}$ .

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