MCQ

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

#### Options

(0, 1)

(−1/4, 1/3)

(0, 1/3)

(0, ∞)

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#### Solution

(−1/4, 1/3)

P(A) = (1 − 3*p*)/2

P(B) = (1 + 4*p*)/3

P(C) = (1 + *p*)/6

The events are mutually exclusive and exhaustive.

∴ P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 1

⇒\[0 \leq P\left( A \right) \leq 1\]

\[0 \leq P\left( B \right) \leq 1\]

\[0 \leq P\left ( C \right) \leq 1\]

⇒ \[0 \leq \frac{1 - 3p}{2} \leq 1\]

\[0 \leq \frac{1 + 4p}{3} \leq 1\]

\[0 \leq \frac{1 + p}{6} \leq 1\]

\[\Rightarrow - 1/3 \leq p \leq \frac{1}{3}\] , ...(i)

\[\frac{- 1}{4} \leq p \leq \frac{1}{2}\] ...(ii)

and \[- 1 \leq p \leq 5\] ...(iii)

The common solution of (i), (ii) and (iii) is\[- 1/4 \leq p \leq 1/3\]

and \[- 1 \leq p \leq 5\] ...(iii)

The common solution of (i), (ii) and (iii) is\[- 1/4 \leq p \leq 1/3\]

∴ The set values of p are (- 1/4 , 1/3)

Concept: Event - Mutually Exclusive Events

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