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
Correct Answer: D
Solution :
[d] 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. |
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