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If $A$ and $B$ are events such that $P(A \cup B)=\frac{5}{6}, P(\bar{A})=\frac{1}{4}$ and $P(B)=\frac{1}{3}$, then $A$ and $B$ are
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Verified Answer
The correct answer is:
independent events
We have,
$$
\begin{array}{rrr}
& P(A \cup B)=\frac{5}{6}, P(\bar{A})=\frac{1}{4}, P(B)=\frac{1}{3} \\
\therefore \quad & P(A)=1-P(\bar{A})=1-\frac{1}{4}=\frac{3}{4} \\
& \text { Also, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\Rightarrow \quad & \frac{5}{6}=\frac{3}{4}+\frac{1}{3}-P(A \cap B) \\
\Rightarrow \quad & P(A \cap B)=\frac{1}{4} \\
\text { and } P(A) \cdot P(B)=\frac{3}{4} \times \frac{1}{3}=\frac{1}{4} \\
\therefore & \quad P(A \cap B)=P(A) \cdot P(B)
\end{array}
$$
Hence, $A$ and $B$ are independent events.
$$
\begin{array}{rrr}
& P(A \cup B)=\frac{5}{6}, P(\bar{A})=\frac{1}{4}, P(B)=\frac{1}{3} \\
\therefore \quad & P(A)=1-P(\bar{A})=1-\frac{1}{4}=\frac{3}{4} \\
& \text { Also, } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
\Rightarrow \quad & \frac{5}{6}=\frac{3}{4}+\frac{1}{3}-P(A \cap B) \\
\Rightarrow \quad & P(A \cap B)=\frac{1}{4} \\
\text { and } P(A) \cdot P(B)=\frac{3}{4} \times \frac{1}{3}=\frac{1}{4} \\
\therefore & \quad P(A \cap B)=P(A) \cdot P(B)
\end{array}
$$
Hence, $A$ and $B$ are independent events.
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