If frac{1 - 3p}{2}, frac{1 + 4p}{3} and frac{ | vedclass

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(B) Since the events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.$\frac{1 - 3p}{2} + \frac{1 + 4p}{3} + \frac{1 +

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$[ - \frac{1}{4}, \frac{1}{3} ]$

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(B) Since the events are mutually exclusive and exhaustive,the sum of their probabilities must be $1$.$\frac{1 - 3p}{2} + \frac{1 + 4p}{3} + \frac{1 + p}{6} = 1$Multiplying by $6$,we get $3(1 - 3p) + 2(1 + 4p) + (1 + p) = 6$.$3 - 9p + 2 + 8p + 1 + p = 6$$6 = 6$,which is always true for any $p$. However,each probability must lie in the interval $[0, 1]$.$1) \ 0 \le \frac{1 - 3p}{2} \le 1 \Rightarrow 0 \le 1 - 3p \le 2 \Rightarrow -1 \le -3p \le 1 \Rightarrow -\frac{1}{3} \le p \le \frac{1}{3}$.$2) \ 0 \le \frac{1 + 4p}{3} \le 1 \Rightarrow 0 \le 1 + 4p \le 3 \Rightarrow -1 \le 4p \le 2 \Rightarrow -\frac{1}{4} \le p \le \frac{1}{2}$.$3) \ 0 \le \frac{1 + p}{6} \le 1 \Rightarrow 0 \le 1 + p \le 6 \Rightarrow -1 \le p \le 5$.Taking the intersection of all these intervals: $[-\frac{1}{3}, \frac{1}{3}] \cap [-\frac{1}{4}, \frac{1}{2}] \cap [-1, 5] = [-\frac{1}{4}, \frac{1}{3}]$.

If $\frac{1 - 3p}{2}, \frac{1 + 4p}{3}$ and $\frac{1 + p}{6}$ are the probabilities of three mutually exclusive and exhaustive events,then the set of all values of $p$ is

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