Question 1180354
<pre>Pay no attention to her gripes.</pre> 
For two events A and B, P(A) = 0.3 and P(B)=0.2. (a) If A and B are independent, then 
P(A|B) = P(A∩B) =
P(A∪B) =<pre>
Venn diagrams make things easier to understand than formulas alone.

Draw a Venn diagram with sets A and B, with r,s,t,u representing the
probabilities of the four regions:

{{{drawing(450,300,-4,4,-2,4.8,
rectangle(-4,-1.6,4,4.4), locate(-2,1.8,r),locate(1.5,1.7,t),
locate(-3.7,-1,u),
 locate(-3.6,2.5,A), locate(-.1,1.8,s),
red(circle(-sqrt(2),sqrt(2),2)),
red(circle(-sqrt(2),sqrt(2),1.95)),
red(circle(-sqrt(2),sqrt(2),1.975)),
blue(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,B)
 )}}}

P(A) = 0.3 = r+s
P(B) = 0.2 = s+t

Since they are independent,

P(A∩B) = s = P(A)P(B) = (0.3)(0.2) = 0.06    <b><--ANSWER to P(A∩B)</b>

So since

s = 0.06 and
r+s = 0.3, substitution gives

r+0.06 = 0.3
r = 0.3-0.06
r = 0.24

And since

s = 0.06 and
s+t = 0.2
0.06+t = 0.2
t = 0.14

P(A∪B) = r+s+t = 0.24+0.06+0.14 = 0.44   <b><--ANSWER to P(A∪B)</b>

{{{drawing(450,300,-4,4,-2,4.8,
rectangle(-4,-1.6,4,4.4), locate(-2,1.8,r=.24),locate(1.5,1.7,t=0.14),
locate(-3.7,-1,u=0.56),
 locate(-3.6,2.5,A), locate(-.5,1.8,s=0.06),
red(circle(-sqrt(2),sqrt(2),2)),
red(circle(-sqrt(2),sqrt(2),1.95)),
red(circle(-sqrt(2),sqrt(2),1.975)),
blue(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,B)
 )}}}

Note: We weren't asked for P(A'∩B') = u, but we could have
found it because r+s+t+u = 1, so 

 0.24+0.06+0.14+u = 1
           0.44+u = 1
                u = 0.56 

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

(b) If A and B are dependent and P(A|B) = 0.25,
then P(B|A) =
P(A∩B) =

{{{drawing(450,300,-4,4,-2,4.8,
rectangle(-4,-1.6,4,4.4), locate(-2,1.8,r),locate(1.5,1.7,t),
locate(-3.7,-1,u),
 locate(-3.6,2.5,A), locate(-.1,1.8,s),
red(circle(-sqrt(2),sqrt(2),2)),
red(circle(-sqrt(2),sqrt(2),1.95)),
red(circle(-sqrt(2),sqrt(2),1.975)),
blue(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,B)
 )}}}

P(A|B) = P(A∩B)/P(B) = s/0.2 = 0.25
                           s = 0.25(0.2)
                           s = 0.05

P(B|A) = P(B∩A)/P(A) = P(A∩B)/P(A) = s/0.2 = 0.05/0.3 = 5/30 = 1/6
                     
(We could have just divided 0.05 by 0.3 and gotten 0.1666666...,
but repeating decimals are a hassle, so I multiplied top and bottom
by 100 and got the fraction 5/30 which reduced to 1/6.

We aren't asked to, but we could finish out the Venn diagram for (b) if we
like:

{{{drawing(450,300,-4,4,-2,4.8,
rectangle(-4,-1.6,4,4.4), locate(-2,1.8,r=.25),locate(1.5,1.7,t=0.15),
locate(-3.7,-1,u=0.55),
 locate(-3.6,2.5,A), locate(-.5,1.8,s=0.06),
red(circle(-sqrt(2),sqrt(2),2)),
red(circle(-sqrt(2),sqrt(2),1.95)),
red(circle(-sqrt(2),sqrt(2),1.975)),
blue(circle(sqrt(2),sqrt(2),2),circle(sqrt(2),sqrt(2),1.95),circle(sqrt(2),sqrt(2),1.975)),
locate(3.4,2.5,B)
 )}}}


Edwin</pre>