Let E 1 , E 2 , E 3 be three mutually exclusive events such that P E 1 = 2 + 3 p 6 , P E 2 = 2 - p 8 and P E…

We know that,

$0\le P\left(E_i\right)\le 1$for $i=1,2,3$

$0\le P\left(E_1\right)\le 1$

$\Rightarrow 0\le \frac{2+3P}{6}\le 1$

$\Rightarrow P\in \left[-\frac{2}{3},\frac{4}{3}\right] ........\left(i\right)$

Now for $0\le P\left(E_2\right)\le 1$

$\Rightarrow 0\le \frac{2-P}{8}\le 1$

$\Rightarrow P\in \left[-6,2\right]........\left(ii\right)$

Now for $0\le P\left(E_3\right)\le 1$

$\Rightarrow 0\le \frac{1-P}{2}\le 1$

$\Rightarrow P\in \left[-1,1\right].......\left(iii\right)$

Also, $E_1$ and $E_2$ and $E_3$ are mutually exclusive

$P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)\le 1$

$\Rightarrow 0\le \frac{13}{12}-\frac{P}{8}\le 1$

$\Rightarrow P\in \left[\frac{2}{3},\frac{26}{3}\right]...........\left(iv\right)$

Now taking intersection of all we get,

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

So maximum value $p_1=1$ and minimum value $p_2=\frac{2}{3}$

So, $p_1+p_2=\frac{5}{3}$