Question 1203558: If A and B are events, show that:
(a) P(A n B') = P(A) - P(A n B)
(b) P(A u B) = i - P(A' n B')
note:
A' and B' is complement
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Problem 1
Draw a rectangle to represent the universal set (aka sample space). Inside the rectangle place partially overlapping circles A and B.
Shade circle A.
If you are in region A, but outside B, then you are in region A n B' where the "n" refers to set intersection.
If you are in A and also in B, then you are located in set A n B.
This is where the circles overlap.
If we started with set A, and kicked out members of set A n B, then we're left with members of set A n B' only.
This is an informal way to prove that
P(A n B') = P(A) - P(A n B)
Another way to prove this would be to look at the law of total probability.
P(A) = P(A n B') + P(A n B)
which rearranges to
P(A n B') = P(A) - P(A n B)
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Problem 2
P(A u B) = 1 - P( (A u B)' )
P(A u B) = 1 - P( A' n B' )
I used the complement rule P(A) = 1-P(A') on the first line.
Then I used De Morgan's Law on the second line.
In terms of a Venn Diagram, region A u B is anywhere in the two circles.
P(A' n B') is the probability of landing outside both circles.
1 - P(A' n B') is the complement of this, and is the probability of landing in region A u B.
Answer by ikleyn(52824)  (Show Source):
You can put this solution on YOUR website! .
If A and B are events, show that:
(a) P(A n B') = P(A) - P(A n B)
(b) P(A u B) = i - P(A' n B')
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(a) P(A n B') in the left side is the probability of events that belong A and B' ;
in other words, it is the probability of events that belong A but do not belong B.
P(A) - P(A n B) in the right side is the probability of events that belong to A but do not belong to B.
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| So, in both sides we have equal quantities. |
| Thus statement (a) is proved. |
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(b) P(A U B) in the left side is the probability of events that belong A or B.
P(A' n B') in the right side is the probability of events that do belong NEITHER A NOR B.
At this point, it is clear that probabilities P(A U B) and P(A' n B') are supplementary.
It is exactly what the statement (b) means.
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| Thus statement (b) is proved. |
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Solved.
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