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 ?

  • 2
    $\begingroup$ The probability of any event is nonnegative; the sum of the probabilities of mutually exclusive events is at most 1. $\endgroup$ Dec 17, 2013 at 18:42

2 Answers 2


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 $\left[-1/3,1/2\right]$.

  • $\begingroup$ If $p<-1/3$, the probability of the first event appears to become negative. Bzzt :-) $\endgroup$ Dec 17, 2013 at 18:58
  • $\begingroup$ Thanks, so we got to check every probability individually. I'll edit it accordingly. $\endgroup$
    – Ian Mateus
    Dec 17, 2013 at 19:00

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.

  • 1
    $\begingroup$ There is a typo in your answer, it should read $\frac{1+3p}{3}$. $\endgroup$
    – Ian Mateus
    Dec 17, 2013 at 18:54
  • $\begingroup$ Nice, thanks a lot. $\endgroup$
    – Asinomás
    Dec 17, 2013 at 18:58

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