Probability Question 268
Question: If A and B are two events. The probability that at most one of A, B occurs, is
Options:
A) $ 1-P(A\cap B) $
B) $ P(\bar{A})+P(\bar{B})-P(\bar{A}\cap \bar{B}) $
C) $ P(\bar{A})+P(\bar{B})+P(A\cup B)-1 $
D) All of these
Show Answer
Answer:
Correct Answer: D
Solution:
Required probability $ =P(\bar{A}\cup \bar{B})=P(\overline{A\cap B}) $
$ =1-P(A\cap B) $
Again, $ P(\bar{A}\cup \bar{B})=P(\bar{A})+P(\bar{B})-P(\bar{A}\cap \bar{B}) $
[By add. Theorem]
Again, $ P(\bar{A}\cup \bar{B})=P(\bar{A})+P(\bar{B})-P(\bar{A}\cap \bar{B}) $
$ =P(\bar{A})+P(\bar{B})-P(\overline{A\cup B}) $
$ =P(\bar{A})+P(\bar{B})-{1-P(A\cup B)} $
$ =P(\bar{A})+P(\bar{B})+P(A\cup B)-1 $
Finally,$ P(\bar{A}\cup \bar{B})=P[(A\cap \bar{B})\cup (\bar{A}\cap B)\cup (\bar{A}\cap \bar{B})] $
$ =P(A\cap \bar{B})+P(\bar{A}\cap B)+P(\bar{A}\cap \bar{B}) $
[ $ \because A\cap \bar{B},\bar{A}\cap B $ and $ \bar{A}\cap \bar{B} $ are mutually exclusive events]
So, alternative [d] is the correct answer.
BETA
