Question 215380
 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 )

<font size = 8 color = "red"><b>Edwin'solution:</font></b>
<pre><font size = 4 color="indigo"><b>
First draw a big rectangle for the universal set:
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4) )}}}
 
Next draw a circle to and label it E:
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), 
 locate(-3.5,2.5,E),
circle(-sqrt(2),sqrt(2),2) )}}}
 
Next draw a circle overlapping the first circle and
label it F. 
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), 
 locate(-3.5,2.5,E),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2)
 )}}}
 
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: 
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), 
 locate(-3.5,2.5,E), locate(-.4,1.8,0.15),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2)
 )}}}
 
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.
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), locate(-2,1.8,0.15),
 locate(-3.5,2.5,E), locate(-.4,1.8,0.15),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2)
 )}}} 
 
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.
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), locate(-2,1.8,0.15), locate(1.5,1.7,0.3),
locate(-3.5,2.5,E), locate(-.4,1.8,0.15),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2)
 )}}} 
 
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:
 
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), locate(-2,1.8,0.15), locate(1.5,1.7,0.3),
locate(-3.5,2.5,E), locate(-.4,1.8,0.15),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2), locate(-3.5,-.7,0.4)
 )}}}

To find

P(E|F<sup>C</sup>)

F<sup>C</sup> means that we are eliminating circle F,
we cross out the probabilities in both parts of F,
and we hove this:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), locate(-2,1.8,0.15), locate(1.5,1.7,cross(0.3)),
locate(-3.5,2.5,E), locate(-.4,1.8,cross(0.15)),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2), locate(-3.5,-.7,0.4)
 )}}}


To find 

P(E|F<sup>C</sup>) 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|F<sup>C</sup>) = {{{0.15/(.4+.15)=0.15/0.55=15/55=3/11}}}

There is another way to find

P(E|F<sup>C</sup>) by the formula

{{{"P(A|B)"=P(A_B)/P(B)}}}

{{{"P(E|F^C)"=P(E_F^C)/P(F^C)=(.15)/(.4+.15)=.15/.55=15/55=3/11}}}

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

To find

P(F|E<sup>C</sup>)

E<sup>C</sup> means that we are eliminating circle E,
we cross out the probabilities in both parts of E,
and we hove this:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-1.2,4,4), locate(-2,1.8,cross(0.15)), locate(1.5,1.7,0.3),
locate(-3.5,2.5,E), locate(-.4,1.8,cross(0.15)),
circle(-sqrt(2),sqrt(2),2),locate(3.5,2.5,F),
circle(sqrt(2),sqrt(2),2), locate(-3.5,-.7,0.4)
 )}}}


To find 

P(F|E<sup>C</sup>) 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|E<sup>C</sup>) = {{{0.3/(0.4+0.3)=0.3/0.7=3/7}}}

The other way to find

P(F|E<sup>C</sup>) is again by the formula

{{{"P(A|B)"=P(A_B)/P(B)}}}

{{{"P(F|E^C)"=P(F_E^C)/P(E^C)=(0.3)/(0.4+0.3)=0.3/0.7=3/7}}}

Edwin</pre>