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If (1+3p)/3 ,(1-p)/4 and (1-2p)/2 are the probabilities of mutually exclusive events, then the set of all values of p is?

My solution- Let the events be A,B,C respectively then 0≤ P(A)+P(B)+P(C)≤1 I could get the minimum value but I am having problem with the maximum. Is it necessary to use 0≤P(A)≤1 and so on for each event ? What's wrong with my solution?

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    Conditions: $P(A), P(B), P(C)\geq0$ and $P(A)+P(B)+P(C)\leq1$ – 2017-02-12
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    The conditions (1+3p)/3>=0, (1-p)/4>=0, (1-2p)/2>=0 and (1+3p)/3+(1-p)/4+(1-2p)/2<=1 are necessary and sufficient. – 2017-02-12
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    I have given an answer. Hope it helped you. – 2017-02-12

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Hint:

Since, $\frac{1+3p}{3}, \frac{1-p}{4}, \frac{1-2p}{2}$ are the probabilities of the three events, we must have

$$\begin{cases} 0 \leq \displaystyle \frac{1+3p}{3}\leq 1 \\ 0 \leq \displaystyle \frac{1-p}{4} \leq 1 \\ 0 \leq \displaystyle \frac{1-2p}{2} \leq 1 \\ \end{cases} \tag{1}$$

Since the three events are mutually exclusive, we also have, $$0 \leq \frac{1+3p}{3} + \frac{1-p}{4} + \frac{1-2p}{2} \leq 1 \tag{2}$$

Use both the conditions and get the common interval. Hope it helps.