Probability Question 178
Question: Let $ 0<P(A)<1 $ , $ 0<P(B)<1 $ and $ P(A\cup B)= $
$ P(A)+P(B)-P(A)P(B). $ Then
[IIT 1995]
Options:
A) $ P(B/A)=P(B)-P(A) $
B) $ P(A^{c}\cup B^{c})=P(A^{c})+P(B^{c}) $
C) $ P{{(A\cup B)}^{c}}=P(A^{c})P(B^{c}) $
D) $ P(A/B)=P(A) $
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ P(A\cap B)=P(A)P(B) $
It means $ A $ and $ B $ are independent events so $ A^{c} $ and $ B^{c} $ will also be independent.
Hence $ P{{(A\cup B)}^{c}}=P(A^{c}\cap B^{c})=P(A^{c})P(B^{c}) $ (Demorgan’s law) As $ A $ is independent of $ B, $ hence $ P(A/B)=P(A) $ , $ {\because P(A\cap B)=P(B)P(A/B)} $ .
BETA
