If frac{1+3P}{3} and frac{1-2P}{2} are probab | vedclass

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Question

(A) For any event $E$,the probability $P(E)$ must satisfy $0 \leq P(E) \leq 1$. For mutually exclusive events $A$ and $B$,$P(A \cup B) = P(A) + P(B) \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For any event $E$,the probability $P(E)$ must satisfy $0 \leq P(E) \leq 1$. For mutually exclusive events $A$ and $B$,$P(A \cup B) = P(A) + P(B) \leq 1$. Step $1$: Constraints on individual probabilities: $0 \leq \frac{1+3P}{3} \leq 1$ $\Rightarrow 0 \leq 1+3P \leq 3$ $\Rightarrow -1 \leq 3P \leq 2$ $\Rightarrow -\frac{1}{3} \leq P \leq \frac{2}{3}$. $0 \leq \frac{1-2P}{2} \leq 1$ $\Rightarrow 0 \leq 1-2P \leq 2$ $\Rightarrow -1 \leq -2P \leq 1$ $\Rightarrow -\frac{1}{2} \leq P \leq \frac{1}{2}$. Step $2$: Constraint for mutually exclusive events: $P(A) + P(B) \leq 1 $ $\Rightarrow \frac{1+3P}{3} + \frac{1-2P}{2} \leq 1 $ $\Rightarrow \frac{2(1+3P) + 3(1-2P)}{6} \leq 1 $ $\Rightarrow 2+6P+3-6P \leq 6 $ $\Rightarrow 5 \leq 6$. This is always true. Step $3$: Intersection of all intervals: $P \in [-\frac{1}{3}, \frac{2}{3}] \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{3}, \frac{1}{2}]$.

If $\frac{1+3P}{3}$ and $\frac{1-2P}{2}$ are probabilities of two mutually exclusive events,then $P$ lies in the interval.

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