Probability 2 Question 6
6. The probability that at least one of the events $A$ and $B$ occurs is 0.6 . If $A$ and $B$ occur simultaneously with probability 0.2 , then $P(\bar{A})+P(\bar{B})$ is equal to $(1987,2 M)$
(a) 0.4
(b) 0.8
(c) 1.2
(d) 1.4
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Answer:
Correct Answer: 6. $(a, c)$
Solution:
- Given, $P(A \cup B)=0.6, P(A \cap B)=0.2$
$$ \begin{aligned} \therefore \quad P(\bar{A}) & +P(\bar{B})=[1-P(A)]+[1-P(B)] \\ & =2-[P(A)+P(B)] \\ & =2-[P(A \cup B)+P(A \cap B)] \\ & =2-[0.6+0.2]=1.2 \end{aligned} $$
