If the events A and B are mutually exclusive | vedclass

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Question

(C) For any event $E$,the probability $P(E)$ must satisfy $0 \le P(E) \le 1$. For event $A$: $0 \le \frac{3x+1}{3} \le 1 \Rightarrow 0 \le 3x+1 \le 3

Vedclass · Accepted Answer

$[-\frac{1}{3}, \frac{5}{9}]$

Vedclass · Answer

(C) For any event $E$,the probability $P(E)$ must satisfy $0 \le P(E) \le 1$. For event $A$: $0 \le \frac{3x+1}{3} \le 1 \Rightarrow 0 \le 3x+1 \le 3 \Rightarrow -1 \le 3x \le 2 \Rightarrow -\frac{1}{3} \le x \le \frac{2}{3}$. For event $B$: $0 \le \frac{1-x}{4} \le 1 \Rightarrow 0 \le 1-x \le 4 \Rightarrow -1 \le -x \le 3 \Rightarrow -3 \le x \le 1$. Since $A$ and $B$ are mutually exclusive,$P(A \cup B) = P(A) + P(B) \le 1$. $\frac{3x+1}{3} + \frac{1-x}{4} \le 1 \Rightarrow \frac{4(3x+1) + 3(1-x)}{12} \le 1 \Rightarrow 12x + 4 + 3 - 3x \le 12 \Rightarrow 9x + 7 \le 12 \Rightarrow 9x \le 5 \Rightarrow x \le \frac{5}{9}$. Combining all conditions: $x \ge -\frac{1}{3}$,$x \le \frac{2}{3}$,$x \ge -3$,$x \le 1$,and $x \le \frac{5}{9}$. The intersection of these intervals is $[-\frac{1}{3}, \frac{5}{9}]$.

If the events $A$ and $B$ are mutually exclusive events such that $P(A) = \frac{3x + 1}{3}$ and $P(B) = \frac{1 - x}{4}$,then the set of possible values of $x$ lies in the interval

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