Question

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 $P (\overline{ A })=\frac{1}{4}$, where $\overline{ A }$ stands for complement of event $A$. Then $A$ and $B$ are

• A. mutually exclusive and independent.
• B. Independent but not equally likely.
• C. equally likely but not independent.
• D. equally likely and mutually exclusive.

Answer 1

Answer: (B)

$P ( A \cap B )=\frac{1}{4} \neq 0$
Therefore, A and B are not mutually exclusive events
Now,
$\text { Now, } P(\overline{A \cup B})=\frac{1}{6} $
$1 P(A \cup B)=\frac{1}{6} $
$P(A \cup B) =\frac{5}{6} $
$P(A \cup B) P(A)+P(B)-P(A \cap B) $
$\frac{5}{6}=\frac{3}{4}+P(B) -\frac{1}{4} $
$P(B) =\frac{5}{6}+\frac{1}{4}-\frac{3}{4}=\frac{1}{3}$
$P(A \cap B) =\frac{1}{4} $
$P(A) P(B)=\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}$
So, $A$ and $B$ are independent But $P ( A ) \neq P ( B ) \Rightarrow A$ and $B$ are not equally likely.