E_1,E_2,E_3 mutually exclusive: P(E_1)=2+3p6, P(E_2)=2-p8, P(E_3)=1-p2.

Question

$E_1,E_2,E_3$ mutually exclusive: $P(E_1)=\dfrac{2+3p}{6}$, $P(E_2)=\dfrac{2-p}{8}$, $P(E_3)=\dfrac{1-p}{2}$. If max/min values of $p$ are $p_1,p_2$, find $p_1+p_2$.

Accepted Answer

Requiring each probability in $[0,1]$ and their sum $\le1$: $p\in\left[\dfrac23,1\right]$ (intersection of all constraints).

$$p_1+p_2 = 1+\frac23=\frac53$$

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