Mathematics › Probability › Addition and Subtraction Theorems
Events $A, B$ and $C$ are mutually exclusive events such that $P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}$ and…
Events $A, B$ and $C$ are mutually exclusive events such that $P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}$ and $P(C)=\frac{1-2 x}{2}$. The set of possible values of $x$ are in the interval
$\left[\frac{1}{3}, \frac{1}{2}\right]$ $\left[\frac{1}{3}, \frac{2}{3}\right]$ $\left[\frac{1}{3}, \frac{13}{3}\right]$ $[0,1]$ Solution
$A, B, C$ are mutually exclusive.
$
\begin{aligned}
& P(A \cap B \cap C)=0, P(A \cap B)=0, \\
& \quad P(B \cap C)=P(C \cap A)=0 \\
& \quad P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4}, P(C)=\frac{1-2 x}{2} \\
& \therefore P(A), P(B), P(C) \in[0,1]
\end{aligned}
$
Now,
\begin{array}{lll}
\hline 0 \leq \frac{3 x+1}{3} \leq 1 & 0 \leq \frac{1-x}{4} \leq 1 & 0 \leq \frac{1-2 x}{2} \leq 1 \\
\hline 0 \leq 3 x+1 \leq 3 & 0 \leq 1-x \leq 4 & 0 \leq 1-2 x \leq 2 \\
\hline-1 \leq 3 x \leq 2 & -1 \leq-x \leq 3 & -1 \leq-2 x \leq 1 \\
\hline-\frac{1}{3} \leq x \leq \frac{2}{3} & -3 \leq x \leq 1 & -\frac{1}{2} \leq x \leq \frac{1}{2} \\
\hline
\end{array}
$\begin{aligned} & P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B) \\ &-P(B \cap C)-P(C \cap A)+P(A \cap B \cap C) \\ & \Rightarrow P(A \cup B \cup C)=P(A)+P(B)+P(C) \\ &=\frac{3 x+1}{3}+\frac{1-x}{4}+\frac{1-2 x}{2} \\ &=\frac{12 x+4+3-3 x+6-12 x}{12}=\frac{13-3 x}{12}\end{aligned}$
$\begin{aligned} \therefore 0 & \leq \frac{13-3 x}{12} \leq 1 \\ 0 & \leq 13-3 x \leq 12 \\ -13 \leq-3 x & \leq-1 \\ \frac{1}{3} \leq x & \leq \frac{13}{3}\end{aligned}$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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