Probability Ques 43
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: 43.( c)
Solution:
Formula:
Set theoretical notation of probability and some important results:
- 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} $
BETA
