If are probabilities of three mutually exclusive events, then

Engineering

Mathematics

Conditional Probability

Question

If $\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}$ are probabilities of three mutually exclusive events, then

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

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

$\frac{1}{6} \leq p \leq \frac{1}{2}$

None of these

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As $\frac{1 + 4 p}{4}, \frac{1 - p}{2}, \frac{1 - 2 p}{2}$ are probabilities of three mutually exclusive events, we must have

0 \leq \frac{1 + 4 p}{4} \leq 1, 0 \leq \frac{1 - p}{2} \leq 1, 0 \leq \frac{1 - 2 p}{2} \leq 1

and

0 \leq \frac{1 + 4 p}{4} + \frac{1 - p}{2} + \frac{1 - 2 p}{2} \leq 1

\Rightarrow - \frac{1}{4} \leq p \leq \frac{3}{4}, - 1 \leq p \leq 1, - \frac{1}{2} \leq p \leq \frac{1}{2}

and

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

\Rightarrow max \{- \frac{1}{4}, - 1, - \frac{1}{2}, \frac{1}{2}\} < p \leq min \{\frac{3}{4}, 1, \frac{1}{2}, \frac{5}{2}\}

\Rightarrow \frac{1}{2} \leq p \leq \frac{1}{2} \Rightarrow p = \frac{1}{2}