Let $$E_1, E_2, E_3$$ be three mutually exclusive events such that $$P(E_1) = \dfrac{2+3p}{6}$$, $$P(E_2) = \dfrac{2-p}{8}$$ and $$P(E_3) = \dfrac{1-p}{2}$$. If the maximum and minimum values of $$p$$ are $$p_1$$ and $$p_2$$ then $$(p_1 + p_2)$$ is equal to:

We need to find the range of $$p$$ such that $$E_1, E_2, E_3$$ are valid mutually exclusive events.

First, impose the non-negativity conditions. Each probability must be $$\ge 0$$:

$$P(E_1) = \dfrac{2 + 3p}{6} \ge 0 \implies p \ge -\dfrac{2}{3}$$

$$P(E_2) = \dfrac{2 - p}{8} \ge 0 \implies p \le 2$$

$$P(E_3) = \dfrac{1 - p}{2} \ge 0 \implies p \le 1$$

Next, consider the sum condition since the events are mutually exclusive:

$$\dfrac{2 + 3p}{6} + \dfrac{2 - p}{8} + \dfrac{1 - p}{2} \le 1$$

Taking LCM = 24:

$$4(2 + 3p) + 3(2 - p) + 12(1 - p) \le 24$$

$$8 + 12p + 6 - 3p + 12 - 12p \le 24$$

$$26 - 3p \le 24$$

$$p \ge \dfrac{2}{3}$$

From the non-negativity conditions we have $$-\dfrac{2}{3} \le p \le 1$$ and from the sum condition $$p \ge \dfrac{2}{3}$$. This gives $$\dfrac{2}{3} \le p \le 1$$.

Finally, to find $$p_1 + p_2$$, note that the maximum value $$p_1 = 1$$ and minimum value $$p_2 = \dfrac{2}{3}$$. Hence $$p_1 + p_2 = 1 + \dfrac{2}{3} = \dfrac{5}{3}$$.

The correct answer is Option B: $$\dfrac{5}{3}$$.