If frac{1 + 3p}{3}, frac{1 - p}{4} and frac{1 | vedclass

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(A) Since $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three events,each probability must lie in the interval $

Vedclass · Accepted Answer

$\frac{1}{3} \le p \le \frac{1}{2}$

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(A) Since $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three events,each probability must lie in the interval $[0, 1]$.$1$) $0 \le \frac{1 + 3p}{3} \le 1 \Rightarrow 0 \le 1 + 3p \le 3 \Rightarrow -1 \le 3p \le 2 \Rightarrow -\frac{1}{3} \le p \le \frac{2}{3}$$2$) $0 \le \frac{1 - p}{4} \le 1 \Rightarrow 0 \le 1 - p \le 4 \Rightarrow -1 \le -p \le 3 \Rightarrow -3 \le p \le 1$$3$) $0 \le \frac{1 - 2p}{2} \le 1 \Rightarrow 0 \le 1 - 2p \le 2 \Rightarrow -1 \le -2p \le 1 \Rightarrow -\frac{1}{2} \le p \le \frac{1}{2}$For mutually exclusive events,the sum of probabilities must be $\le 1$:$\frac{1 + 3p}{3} + \frac{1 - p}{4} + \frac{1 - 2p}{2} \le 1$Multiply by $12$: $4(1 + 3p) + 3(1 - p) + 6(1 - 2p) \le 12$$4 + 12p + 3 - 3p + 6 - 12p \le 12$$13 - 3p \le 12 \Rightarrow -3p \le -1 \Rightarrow p \ge \frac{1}{3}$Taking the intersection of all conditions:$p \in [-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] \cap [\frac{1}{3}, \infty)$Intersection is $\frac{1}{3} \le p \le \frac{1}{2}$.

If $\frac{1 + 3p}{3}, \frac{1 - p}{4}$ and $\frac{1 - 2p}{2}$ are the probabilities of three mutually exclusive events,then the set of all values of $p$ is

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