Question

If $$\displaystyle \frac{1+3p}{3}, \frac{1-p}{4}$$ and $$\dfrac{1-2p}{2}$$ are the probabilities of the three mutually exclusive events, then $$p \in$$

A
[0,1]
B
[0,12]
C
[13,1]
D
[13,12]

Solution

The correct option is D $$\left[\dfrac{1}{3}, \dfrac{1}{2}\right]$$Mathematics

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