A) \[P(B/A)=P(B)-P(A)\]
B) \[P(A'-B')=P(A')-P(B')\]
C) \[P(A\cup B)'=P(A).P(B)'\]
D) \[P(A/B)=P(A)-P(B)\]
Correct Answer: C
Solution :
Since, \[P(A\cap B)=P(A).P(B)\] It means A and B are independent events, so A' and B' are also independent. \[\therefore \]\[P(A\cup B)'=P(A'\cap B')\] \[=P(A').P(B')\] Alternative \[P(A\cup B)'=1-P(A\cup B)\] \[=1-[P(A)+P(B)-P(A).P(B)]\] \[=[1-P(A)][1-P(B)]\] \[=P(A)'.P(B)'\]You need to login to perform this action.
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