Events $A,B,C$ 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 $x$ are in the interval:

For mutually exclusive events, we must satisfy:$0\le P(A)\le 1,0\le P(B)\le 1,0\le P(C)\le 1$$P(A)+P(B)+P(C)\le 1$ Conditions from individual probabilities:$0\le \frac{3x+1}{3}\le 1⟹-1\le 3x\le 2⟹-1/3\le x\le 2/3$.$0\le \frac{1-x}{4}\le 1⟹-3\le x\le 1$.$0\le \frac{1-2x}{2}\le 1⟹-1/2\le x\le 1/2$.
Intersection so far: $-1/3\le x\le 1/2$.Condition for sum:
$$\frac{3x+1}{3}+\frac{1-x}{4}+\frac{1-2x}{2}\le 1$$
Multiply by 12:
$$4(3x+1)+3(1-x)+6(1-2x)\le 12$$
$$12x+4+3-3x+6-12x\le 12$$
$$13-3x\le 12⟹1\le 3x⟹x\ge 1/3$$
.
Combining with $x\le 1/2$: Interval is $[1/3,1/2]$.
Checking the answer key provided: (C) is marked but calculations lead to a subset of $[1/3,1/2]$.