The probabilities of three mutually exclusive events If $\cfrac{1+3p}{3}$,$\cfrac{1-p}{4}$ and $\cfrac{1-2p}{6}$ are the probabilities of three mutually exclusive events, then the set of all values of $p$ is:  
$A. \left(\frac{1}{3},\frac{1}{2}\right)$ 
$B. \left[\frac{1}{3},\frac{1}{2}\right]$
$C. \left[\frac{1}{3},\frac{5}{6}\right]$
$D.$ None of these
I know that mutually exclusive events are events which cannot occur at the same time. But then, how do you solve this question with this much knowledge and some elementary probability ?
 A: If they are mutually exclusive, their sum is between $0$ and $1$. Moreover, each probability is between $0$ and $1$, so we have four conditions. Sum everything to get $0\leqslant(5p+9)/12\leqslant 1$. This implies $p\in \left[-9/5, 3/5\right]$. 
The first probability implies $0\leqslant (1+3p)/3 \leqslant 1 \implies p\in\left[-1/3, 2/3\right]$.
The second implies $0\leqslant (1-p)/4 \leqslant 1 \implies p\in\left[-3, 1\right].$
The third implies $0\leqslant (1-2p)/6 \leqslant 1 \implies p\in\left[-5/2, 1/2\right].$
Do the intersections to get $\left[-1/3,1/2\right]$.
A: the sum of the probabilities is $\frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{6}=\frac{3}{4}+\frac{5p}{12}$.
If non of the individual values are negative and the sum is in the interval [0,1] then $p$ is a good value.
for all values to be positive we must have$ p\in[\frac{-1}{3},\frac{1}{2}]$
for the sum to be in the interval we must have $p\in[\frac{-9}{5},\frac{3}{5}]$
so the interval is $[\frac{-1}{3},\frac{1}{2}]$.Because it is the intersection of the two intervals.
Trying with p=0 shows that 0 should be in the interval since all probabilities are between 0 and 1 and their sum is also between 0 and 1.
