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Question

If (1-3p)2,(1+4p)3,(1+p)6 are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is
(a) (0, 1)
(b) (−1/4, 1/3)
(c) (0, 1/3)
(d) (0, ∞)

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Solution

(b) (−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

0PA1, 0PB1, 0PC1

01-3p21, 01+4p31, 01+p61

-1/3p13, ...(i)


-14p12 ...(ii)

and -1p5 ...(iii)
The common solution of (i), (ii) and (iii) is -1/4p1/3.
∴ The set values of p are (-1/4 , 1/3)

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