Q120 / 170 · Manipal (MET) 2016 · Mathematics

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 complement of event $A$. Then, the events $A$ and $B$ are

  • A
    mutually exclusive and independent
  • Correct answer
    B
    independent but not equally likely
  • C
    equally likely but not independent
  • D
    equally likely and mutually exclusive

Solution

Given, $P(\overline{A \cup B})=\frac{1}{6}$ $\begin{array}{lr}\Rightarrow & 1-P(A \cup B)=\frac{1}{6} \\ \Rightarrow & 1-P(A)-P(B)+P(A \cap B)=\frac{1}{6}\end{array}$ $\begin{array}{lr}\Rightarrow & P(\bar{A})-P(B)+P(A \cap B)=\frac{1}{6} \\ \Rightarrow & \frac{1}{4}-P(B)+\frac{1}{4}=\frac{1}{6}\end{array}$ $\begin{aligned} & \Rightarrow \\ & \text { and } \quad P(B)=\frac{1}{3} \\ & \text { an } \quad P(\bar{A})=1-\frac{1}{4}=\frac{3}{4}\end{aligned}$ Since, $P(A \cap B)=P(A) \cdot P(B)$, so the events $A$ and $B$ are independent events but not equally likely.

Hence, the correct answer is option: 'B'.

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