Events A, B and C are mutually exclusive even | vedclass

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(A) Since $A, B, C$ are mutually exclusive,we have $P(A \cap B) = P(B \cap C) = P(C \cap A) = P(A \cap B \cap C) = 0$. For any event $E$,$0 \leq P(E)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Since $A, B, C$ are mutually exclusive,we have $P(A \cap B) = P(B \cap C) = P(C \cap A) = P(A \cap B \cap C) = 0$. For any event $E$,$0 \leq P(E) \leq 1$. $1$. For $P(A) = \frac{3x+1}{3}$: $0 \leq \frac{3x+1}{3} \leq 1 \implies 0 \leq 3x+1 \leq 3 \implies -1 \leq 3x \leq 2 \implies -\frac{1}{3} \leq x \leq \frac{2}{3}$. $2$. For $P(B) = \frac{1-x}{4}$: $0 \leq \frac{1-x}{4} \leq 1 \implies 0 \leq 1-x \leq 4 \implies -1 \leq -x \leq 3 \implies -3 \leq x \leq 1$. $3$. For $P(C) = \frac{1-2x}{2}$: $0 \leq \frac{1-2x}{2} \leq 1 \implies 0 \leq 1-2x \leq 2 \implies -1 \leq -2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2}$. Intersection of these intervals: $[-\frac{1}{3}, \frac{2}{3}] \cap [-3, 1] \cap [-\frac{1}{2}, \frac{1}{2}] = [-\frac{1}{3}, \frac{1}{2}]$. Also,for mutually exclusive events,$P(A \cup B \cup C) = P(A) + P(B) + P(C) \leq 1$. $\frac{3x+1}{3} + \frac{1-x}{4} + \frac{1-2x}{2} \leq 1$ $\frac{4(3x+1) + 3(1-x) + 6(1-2x)}{12} \leq 1$ $\frac{12x+4+3-3x+6-12x}{12} \leq 1 \implies \frac{13-3x}{12} \leq 1 \implies 13-3x \leq 12 \implies -3x \leq -1 \implies x \geq \frac{1}{3}$. Combining $x \geq \frac{1}{3}$ with the intersection $[-\frac{1}{3}, \frac{1}{2}]$,we get $x \in [\frac{1}{3}, \frac{1}{2}]$.

List $I$	List $II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

Events $A, B$ and $C$ are mutually exclusive events such that $P(A)=\frac{3x+1}{3}$,$P(B)=\frac{1-x}{4}$ and $P(C)=\frac{1-2x}{2}$. The set of possible values of $x$ is in the interval

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