Animated Solution for Mathematics - Probability: 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.

In the realm of probability, the likelihood of any event must be strictly confined between $0$ and $1$. For events $A,B,$ and $C$ defined by variable $x$, we must ensure each individual probability satisfies this axiom.

For event $A$, where $P(A)=\frac{3x+1}{3}$, we set: Multiplying by $3$ gives $0\le 3x+1\le 3$. Subtracting $1$ and dividing by $3$, we find the first boundary:

For event $B$, where $P(B)=\frac{1-x}{4}$, we enforce: Multiplying by $4$ gives $0\le 1-x\le 4$. Subtracting $1$ results in $-1\le -x\le 3$. Multiplying by $-1$ flips the inequality signs, yielding the second boundary:

For event $C$, where $P(C)=\frac{1-2x}{2}$, we set: Multiplying by $2$ gives $0\le 1-2x\le 2$. Subtracting $1$ yields $-1\le -2x\le 1$. Dividing by $-2$ and flipping the signs, we arrive at the third boundary:

Since events $A,B,$ and $C$ are mutually exclusive, the probability of their union is the sum of their individual probabilities. This total sum must also satisfy the axiom $P(A\cup B\cup C)\le 1$.

We perform the summation:

Using the least common multiple of $12$, the inequality becomes:

Expanding the numerators, we observe:

Multiplying by $12$ gives $13-3x\le 12$, which simplifies to $1\le 3x$. Thus, we obtain the final condition:

To find the valid range for $x$, we must find the intersection of all four established conditions: 1. $x\in [-\frac{1}{3},\frac{2}{3}]$ 2. $x\in [-3,1]$ 3. $x\in [-\frac{1}{2},\frac{1}{2}]$ 4. $x\in [\frac{1}{3},\infty )$

By comparing the lower bounds, the maximum is $\frac{1}{3}$. By comparing the upper bounds, the minimum is $\frac{1}{2}$.

Therefore, the valid interval for $x$ is: