SOLUTION: Let P(E) = 0.3, P(F) = 0.45, and P(F E) = 0.15. Draw a Venn diagram and find the conditional probabilities. (a) P(E | FC ) (b) P(F | EC )

First draw a big rectangle for the universal set:
 

 
Next draw a circle to and label it E:
 

 
Next draw a circle overlapping the first circle and
label it F. 
 

 
The overlapping part is the event "E and F", or "E F".
We are given P(F E) = 0.15, so we write "0.15" in the 
region that's shaped like this "()", the overlapping 
part opf the two circles: 
 

 
Now since P(E) = 0.3, and the probability of
being in the overlapping part, shaped like thisa "()"
is 0.15, the probability of being in the rest of circle
E is 0.3-0.15 or 0.15, so we write 0.15 in the left part
of circle E, so that the sum of the probabilities of 
being in the two parts of circle E is 0.3.
 
 
 
Now since P(F) = 0.45, and the probability of
being in the overlapping part, shaped like thisa "()"
is 0.15, the probability of being in the rest of circle
E is 0.45-0.15 or 0.3, so we write 0.15 in the left part
of circle E, so that the sum of the probabilities of 
being in the two parts of circle E is 0.3.
 
 
 
Now we have placed the probabilities of being in each of 3 regions
as 0.15, 0.15, and 0.3.  The probability of all four regions must be
equal to 1.  Therefore since we have 0.15 + 0.15 + .3 = 0.6. Then
if we subtract that from 1 we get 1 - 0.6 = 0.4, and so we write
0.4 in the region inside the big rectangle but outside the two
circles: So 0.4 goes in the rectangle outside both circles.
I'll put those 3 people down on the bottom left side of the
rectangle outside both circles:
 


To find

P(E|FC)

FC means that we are eliminating circle F,
we cross out the probabilities in both parts of F,
and we hove this:




To find 

P(E|FC) we form the ratio of the probability not crossed
out in E to the sum of both probabilities not crossed out, so we have:

P(E|FC) = 

There is another way to find

P(E|FC) by the formula





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

To find

P(F|EC)

EC means that we are eliminating circle E,
we cross out the probabilities in both parts of E,
and we hove this:




To find 

P(F|EC) we form the ratio of the probability not crossed
out in F to the sum of both probabilities not crossed out, so we have:

P(F|EC) = 

The other way to find

P(F|EC) is again by the formula





Edwin