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प्रश्न
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, ∞)
उत्तर
(−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
⇒\[0 \leq P\left( A \right) \leq 1\]
\[0 \leq \frac{1 + 4p}{3} \leq 1\]
\[0 \leq \frac{1 + p}{6} \leq 1\]
\[\Rightarrow - 1/3 \leq p \leq \frac{1}{3}\] , ...(i)
and \[- 1 \leq p \leq 5\] ...(iii)
The common solution of (i), (ii) and (iii) is\[- 1/4 \leq p \leq 1/3\]
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