SOLUTION: P(A)=.37, P(B)=.19, P(A and B)=.13. What is the probability that A doesn't occur or B does?

There are two ways to do this problem. By formulas and by Venn diagram.
I'll show you both ways, because you need to learn to do it both ways.

First, by formulas:

A prime " ' " after an event such as B' is the event that B does not occur.

We'll need three formulas:

1.     P(X or Y) = P(X) + P(Y) - P(X and Y)

2.     P(X') = 1 - P(X)

3.     P(X) = P(X and Y) + P(X and Y')

If we substitute in formula 1, using X=A' and Y=B,

       P(A' or B) = P(A') + P(B) - P(A' and B)
       P(A' or B) = P(A') + .19  - P(A' and B)

So we will need both P(A') and P(A' and B) to complete that. 

We can find P(A') from formula 1, using X=A

       P(A') = 1 - P(A) = 1 - .37 = .63

We can find P(A' and B) from formula 3, using X=B and Y=A.

       P(B) = P(B and A) + P(B and A')
       .19  =     .13    + P(B and A')
       .06  = P(B and A')   <--- that is the same as P(A' and B) 

Now we have what we need to complete

       P(A' or B) = P(A') + .19  - P(A' and B)
       P(A' or B) = .63 + .19   - .06
       P(A' or B) = .76 

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We can also do the problem with a Venn diagram:



P(A)=.37, P(B)=.19, P(A and B)=.13. What is the probability that A doesn't occur or B does?

The red circle is set A and the blue circle is set B.
The region #1 is the part of the set A that is NOT part of the set B 
The region #3 is the part of the set B that is NOT part of the set A
The region #4 is neither part of set A nor part of set B.   

P(A and B)=.13  so we write .13 in region #2



Now since the probability of A, the entire red set, is .37.
Since part of the red set already has .13 in it, since both
parts of set A has to sum to .37, we can subtract .37-.13 = .24
and so we write .24 in the left part of set A, region #1.  
That makes the sum of the probabilities of the two parts of set 
A be .37.



Now since the probability of B, the entire blue set, is .19.
Since part of the blue set already has .13 in it, since both
parts of set B has to sum to .19, we can subtract .19-.13 = .06
and so we write .06 in the right part of set B, region #3.  
That makes the sum of the probabilities of the two parts of set 
B be .19.



Since all four of the regions must have a total probability of 1,
we can add the three probabilities we have found .24+.13+.06 = .43
and subtracting from 1, 1-.43 = .57

So we put .57 for the region #4 which is outside both circles:



Now finally we go through each of the four regions and ask the question:

Does this region represent a case where either A doesn't occur or B does?

We ask that question about region #1 which has probability .24.
Does this region represent a case where either A doesn't occur or B does?
No it does not, because A occurs and B doesn't. So we do not include 
this region.

We ask that question about region #2 which has probability .13.
Does this region represent a case where either A doesn't occur or B does?
Yes it does, because B occurs in region #2. So we do include 
this region.  So we include .13

We ask that question about region #3 which has probability .06.
Does this region represent a case where either A doesn't occur or B does?
Yes it does, because B occurs in region #3. So we do include 
this region.  So we include .06.  So far we have .13 and .06 to add.

We ask that question about region #4 which has probability .57.
Does this region represent a case where either A doesn't occur or B does?
Yes it does, because A does not occur in region #4. So we do include 
this region.  So we include .57.  So the final answer is found by
adding the three probabilities of the regions which passed the test of
representing a case where either A doesn't occur or B does.

.13+.06+.57 = .76

Edwin