Question

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

Answer 1

Let,

$P\left(A\right)=\frac{1+3p}{3},P\left(B\right)=\frac{1-p}{4},P\left(C\right)=\frac{1-2p}{2}$

A,B and C are three mutually exclusive events.

$\therefore P\left(A\right)+P\left(B\right)+P\left(C\right)\le 1$

$\Rightarrow \frac{1+3p}{3}+\frac{1-p}{4}+\frac{1-2p}{2}\le 1\le 1$

$\Rightarrow 4+12p+3-3p+6-12p\le 12$

$\Rightarrow p\le 1/3$................(1)

Also,

$0\le P\left(A\right)\le 1\Rightarrow 0\le \frac{1+3p}{3}\le 1$

$0\le 1+3p\le 3$

$\Rightarrow -\frac{1}{3}\le p\le \frac{2}{3}$............(2)

$0\le P\left(B\right)\le 1\Rightarrow 0\le \frac{1-p}{4}\le 1$

$\Rightarrow 0\le 1-p\le 4$

$\Rightarrow -3\le p\le 1$.................(3)

$0\le P\left(C\right)\le 1\Rightarrow 0\le \frac{1-2p}{2}\le 1$

$\Rightarrow -\frac{1}{2}\le p\le \frac{1}{2}$.................(4)

Combining eq.(1),(2),(3) and (4) ,we get

$\frac{1}{3}\le p\le \frac{1}{2}$