SOLUTION: 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

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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)

-----------------------

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(52832)   (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')
~~~~~~~~~~~~~~~~~~~~~ 

(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.

                  +--------------------------------------------------+
                  |    So, in both sides we have equal quantities.   |
                  |          Thus statement (a) is proved.           |
                  +--------------------------------------------------+



(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.


                  +--------------------------------------------------+
                  |         Thus statement (b) is proved.            |
                  +--------------------------------------------------+

Solved.



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