Let E _ 1 , E _ 2 , E _ 3 be three mutually exclusive events such...

We know that,

$0 \le P\left({E}_{i}\right) \le 1$for $i = 1, 2, 3$

$0 \le P\left({E}_{1}\right) \le 1$

$⇒ 0 \le \dfrac{2 + 3P}{6} \le 1$

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

Now for $0 \le P\left({E}_{2}\right) \le 1$

$⇒ 0 \le \dfrac{2 - P}{8} \le 1$

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

Now for $0 \le P\left({E}_{3}\right) \le 1$

$⇒ 0 \le \dfrac{1 - P}{2} \le 1$

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

Also, ${E}_{1}$ and ${E}_{2}$ and ${E}_{3}$ are mutually exclusive

$P\left({E}_{1}\right) + P\left({E}_{2}\right) + P\left({E}_{3}\right) \le 1$

$⇒ 0 \le \dfrac{13}{12} - \dfrac{P}{8} \le 1$

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

Now taking intersection of all we get,

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

So maximum value ${p}_{1} = 1$ and minimum value ${p}_{2} = \dfrac{2}{3}$

So, ${p}_{1} + {p}_{2} = \dfrac{5}{3}$