A, B and C are mutually exclusive events. If | vedclass

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(C) Since $A, B$ and $C$ are mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C)$.Also,$P(A \cup B \cup C) \leq 1$.Thus,$0 \leq \frac{3

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$\left[ \frac{1}{3}, \frac{1}{2} \right]$

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(C) Since $A, B$ and $C$ are mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C)$.Also,$P(A \cup B \cup C) \leq 1$.Thus,$0 \leq \frac{3x + 1}{3} + \frac{1 - x}{4} + \frac{1 - 2x}{2} \leq 1$.Multiplying by $12$,we get $0 \leq 4(3x + 1) + 3(1 - x) + 6(1 - 2x) \leq 12$.$0 \leq 12x + 4 + 3 - 3x + 6 - 12x \leq 12$.$0 \leq 13 - 3x \leq 12$.Subtracting $13$,we get $-13 \leq -3x \leq -1$.Dividing by $-3$ (reversing the inequality),we get $\frac{1}{3} \leq x \leq \frac{13}{3}$.Additionally,for any event $E$,$0 \leq P(E) \leq 1$:$1$) $0 \leq \frac{3x + 1}{3} \leq 1 \Rightarrow 0 \leq 3x + 1 \leq 3 \Rightarrow -1 \leq 3x \leq 2 \Rightarrow -\frac{1}{3} \leq x \leq \frac{2}{3}$.$2$) $0 \leq \frac{1 - x}{4} \leq 1 \Rightarrow 0 \leq 1 - x \leq 4 \Rightarrow -1 \leq -x \leq 3 \Rightarrow -3 \leq x \leq 1$.$3$) $0 \leq \frac{1 - 2x}{2} \leq 1 \Rightarrow 0 \leq 1 - 2x \leq 2 \Rightarrow -1 \leq -2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}$.Taking the intersection of all conditions: $x \in [\frac{1}{3}, \frac{13}{3}] \cap [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] = [\frac{1}{3}, \frac{1}{2}]$.

$A, B$ and $C$ are mutually exclusive events. If $P(A) = \frac{3x + 1}{3}$,$P(B) = \frac{1 - x}{4}$,and $P(C) = \frac{1 - 2x}{2}$,then the possible values of $x$ lie in the interval ...

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