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

The correct option is D $$\left[\dfrac{1}{3}, \dfrac{1}{2}\right]$$
1917138_1096278_ans_1b0c6769040743aebac47a2bba3754ba.jpg

Mathematics

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