Let E_{1}, E_{2}, E_{3} be three mutually exc | vedclass

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(B) For any event $E_{i}$,$0 \leq P(E_{i}) \leq 1$. For $P(E_{1}) = \frac{2+3p}{6}$,$0 \leq 2+3p \leq 6 \implies -2 \leq 3p \leq 4 \implies -\frac{2}{

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$\frac{5}{3}$

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(B) For any event $E_{i}$,$0 \leq P(E_{i}) \leq 1$. For $P(E_{1}) = \frac{2+3p}{6}$,$0 \leq 2+3p \leq 6 \implies -2 \leq 3p \leq 4 \implies -\frac{2}{3} \leq p \leq \frac{4}{3}$. For $P(E_{2}) = \frac{2-p}{8}$,$0 \leq 2-p \leq 8 \implies -2 \leq -p \leq 6 \implies -6 \leq p \leq 2$. For $P(E_{3}) = \frac{1-p}{2}$,$0 \leq 1-p \leq 2 \implies -1 \leq -p \leq 1 \implies -1 \leq p \leq 1$. Since $E_{1}, E_{2}, E_{3}$ are mutually exclusive,$P(E_{1}) + P(E_{2}) + P(E_{3}) \leq 1$. $\frac{2+3p}{6} + \frac{2-p}{8} + \frac{1-p}{2} \leq 1$. Multiplying by $24$: $4(2+3p) + 3(2-p) + 12(1-p) \leq 24$. $8 + 12p + 6 - 3p + 12 - 12p \leq 24$. $26 - 3p \leq 24 \implies 2 \leq 3p \implies p \geq \frac{2}{3}$. Combining all constraints: $p \in [\frac{2}{3}, 1]$. Thus,$p_{1} = 1$ and $p_{2} = \frac{2}{3}$. $p_{1} + p_{2} = 1 + \frac{2}{3} = \frac{5}{3}$.

Let $E_{1}, E_{2}, E_{3}$ be three mutually exclusive events such that $P(E_{1}) = \frac{2+3p}{6}$,$P(E_{2}) = \frac{2-p}{8}$,and $P(E_{3}) = \frac{1-p}{2}$. If the maximum and minimum values of $p$ are $p_{1}$ and $p_{2}$,then $(p_{1} + p_{2})$ is equal to.

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