Question

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

 

विकल्प

• (0, 1)

•  (−1/4, 1/3)

• (0, 1/3)

• (0, ∞)

   

Answer 1

 (−1/4, 1/3)

P(A) = (1 − 3p)/2
P(B) = (1 + 4p)/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\[- 1/4 \leq p \leq 1/3\]

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