Q125 / 240 · KCET 2026 · Mathematics
Probability of at least one of the events A and B occur is $0.6$. If A and B occur simultaneously with probability $0.2$, then $P(\bar{A}) + P(\bar{B})$ is
- A$1$
- B$0.8$
- C$0.6$
- Correct answerD$1.2$
Solution
Given $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.2$.
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.6 = P(A) + P(B) - 0.2$
$P(A) + P(B) = 0.8$
We need to find $P(\bar{A}) + P(\bar{B})$:
$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B))$
$P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))$
$P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2$
Answer: $1.2$
Using the addition theorem of probability:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$0.6 = P(A) + P(B) - 0.2$
$P(A) + P(B) = 0.8$
We need to find $P(\bar{A}) + P(\bar{B})$:
$P(\bar{A}) + P(\bar{B}) = (1 - P(A)) + (1 - P(B))$
$P(\bar{A}) + P(\bar{B}) = 2 - (P(A) + P(B))$
$P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2$
Answer: $1.2$
Hence, the correct answer is option: 'D'.