Question 1203558
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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.
{{{
drawing(400,200,-5,5,-5,5,
line(-4.5,4.5,4.5,4.5),
line(4.5,4.5,4.5,-4.5),
line(4.5,-4.5,-4.5,-4.5),
line(-4.5,-4.5,-4.5,4.5),
circle(-1,0,2),
circle(1,0,2),
locate(-3,3,"A"),
locate(3,3,"B"),
locate(-2,1,"AnB'"),
locate(0,1,"AnB")
)
}}}


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.
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