The probability that atleast one of the two events $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, evaluate $P(\overline{A}) + P(\overline{B})$.

$1.1$

The correct answer is Option (2) → $1.1$ ##

We know that, $A \cup B$ denotes the occurrence of atleast one of $A$ and $B$ and $A \cap B$ denotes the occurrence of both $A$ and $B$, simultaneously.

Thus, $P(A \cup B) = 0.6 \quad \text{and} \quad P(A \cap B) = 0.3$

Also, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\Rightarrow 0.6 = P(A) + P(B) - 0.3$

$\Rightarrow P(A) + P(B) = 0.9$

$\Rightarrow [1 - P(\overline{A})] + [1 - P(\overline{B})] = 0.9 \quad [∵P(A) = 1 - P(\overline{A}) \text{ and } P(B) = 1 - P(\overline{B})]$

$\Rightarrow P(\overline{A}) + P(\overline{B}) = 2 - 0.9 = 1.1$