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Question

If $$\dfrac {1 + 3p}{3}, \dfrac {1 - p}{4}$$ and $$\dfrac {1 - 2p}{2}$$ are mutually exclusive events. Then, range of $$p$$ is


A
13p12
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B
12p12
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C
13p23
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D
13p25
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Solution

The correct option is A $$\dfrac {1}{3} \leq p \leq \dfrac {1}{2}$$
Since, the probability lies between $$0$$ and $$1$$.
$$0\leq \dfrac {1 + 3p}{3}\leq 1, 0\leq \dfrac {1 - p}{4}\leq 1, 0\leq \dfrac {1 - 2p}{2}\leq 1$$
$$\Rightarrow 0\leq 1 + 3p\leq 3, 0\leq 1 - p\leq 4, 0\leq 1 - 2p \leq 2$$
$$\Rightarrow -\dfrac {1}{3} \leq p\leq \dfrac {2}{3}, -3 \leq p\leq 1, -\dfrac {1}{2} \leq p\leq \dfrac {1}{2} ..... (i)$$
Again, the events are mutually exclusive
$$0\leq \dfrac {1 + 3p}{3} + \dfrac {1 - p}{4} + \dfrac {1 - 2p}{2}\leq 1$$
$$\Rightarrow 0\leq 13 - 3p \leq 12$$
$$\Rightarrow \dfrac {1}{3}\leq p\leq \dfrac {13}{3} .... (ii)$$
From Eqs. (i) and (ii),
$$max\left \{-\dfrac {1}{3}, -3, \dfrac {-1}{2}, \dfrac {1}{3}\right \} \leq p\leq min \left \{\dfrac {2}{3}, 1, \dfrac {1}{2}, \dfrac {13}{3}\right \}$$
$$\Rightarrow \dfrac {1}{3} \leq p\leq \dfrac {1}{2}$$.

Mathematics

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