Probability 3 Question 7
7. Let $A$ and $B$ be two events such that $P(\overline{A \cup B})=\frac{1}{6}$, $P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4}$, where $\bar{A}$ stands for the complement of the event $A$. Then, the events $A$ and $B$ are
(2014 Main)
(a) independent but not equally likely
(b) independent and equally likely
(c) mutually exclusive and independent
(d) equally likely but not independent
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Solution:
- Given, $P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4}, P(\bar{A})=\frac{1}{4}$
$$ \therefore \quad P(A \cup B)=1-P(\overline{A \cup B})=1-\frac{1}{6}=\frac{5}{6} $$
$$ \begin{aligned} \text { and } P(A) & =1-P(\bar{A})=1-\frac{1}{4}=\frac{3}{4} \\ \therefore \quad P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\ \frac{5}{6} & =\frac{3}{4}+P(B)-\frac{1}{4} \\ \Rightarrow \quad P(B) & =\frac{1}{3} \Rightarrow A \text { and } B \text { are not equally likely } \\ P(A \cap B) & =P(A) \cdot P(B)=\frac{1}{4} \end{aligned} $$
So, events are independent.
