Asked by anonymous
With the use of diagrams, show that
P(A′∩B′) = P(A∪B)'
I just need help with understanding how these two equal each other so that i'm able to show it with diagrams. Also don't understand why the apostrophes are there :L, thanks.
P(A′∩B′) = P(A∪B)'
I just need help with understanding how these two equal each other so that i'm able to show it with diagrams. Also don't understand why the apostrophes are there :L, thanks.
Answers
Answered by
Arora
If an event is E, and has probability P(E), then E' is the event of 'E does not happen' and has probability (1-P(E))
Hence,
P(E') = 1 - P(E)
So,
P(A′∩B′) would equal the intersection of 'not A' and 'not B', which can be identified as everything except the two circles in the common Venn Diagram of the representation of sets A and B in a whole.
P(A U B)' would also refer to the same area, because it includes everything that is 'not AUB', that is, everything except the two circles.
Hence,
P(E') = 1 - P(E)
So,
P(A′∩B′) would equal the intersection of 'not A' and 'not B', which can be identified as everything except the two circles in the common Venn Diagram of the representation of sets A and B in a whole.
P(A U B)' would also refer to the same area, because it includes everything that is 'not AUB', that is, everything except the two circles.
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