Question 935211
<pre>
Given:  P(A&#5198;B) = P(A)P(B)
Prove:  P(A&#5198;B') &#8799; P(A)P(B') 

Let P(A&#5198;B')=w
Let P(A&#5198;B)=x
Let P(A'&#5198;B)=y
Let P(A'&#5198;B')=z

Then x+y+w+z = 1

Draw a Venn diagram with the probability of each region written
in each region:

{{{drawing(300,200,-4,4,-2,4.8,rectangle(-4,-1.6,4,4.4), locate(-2,1.8,w),locate(1.5,1.7,y),locate(-3.7,-1,z), locate(-3.6,2.5,A), locate(-.1,1.8,x),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&#5198;B) = P(A)P(B)      given
          x = (w+x)(x+y)    
 x(w+x+y+z) = (w+x)(x+y)    since w+x+y+z = 1
wx+x²+xy+xz = wx+wy+x²+xy   multiplying out
         xz = wy

We want to prove:

    P(A&#5198;B') &#8799; P(A)P(B')
          w &#8799; (w+x)(w+z)
 w(w+x+y+z) &#8799; (w+x)(w+z)    since w+x+y+z = 1  
w²+wx+wy+wz &#8799; w²+wz+wx+xz

So to prove that, we take

         wy = xz            same as xz = wy , proved above

                            and add w²+wx+wz to both sides:

wy+w²+wx+wz = xz+w²+wx+wz   
w²+wx+wy+wz = w²+wz+wx+xz   rearrange to reverse the above steps
 w(w+x+y+z) = (w+x)(w+z)
          w = (w+x)(w+z)    since w+x+y+z = 1
    P(A&#5198;B') = P(A)P(B')     which is what we had to prove.

Edwin</pre>