Question 1008702
<pre>
The other tutor's answer is wrong because he assumed something 
that was not given.  Notice that he wrongly assumed this:
</pre>
>><s>Since P(B) = 0.4, P(A) = 0.6</s><<
<pre>
We CANNOT assume this, for that would only be true if
A and B were given to be mutually exclusive.  But they
are only given to be dependent, not mutually exclusive!
If they were mutually exclusive then P(A|B)=0, not 0.55.
So he was definitely wrong to assume this. 

Since we are not given anything about sets A&#5198;B' or A'&#5198;B', we cannot
find P(A&#5196;B).  However we can find P(A&#5198;B).  Wasn't that really what
you were asked to find, and there is a typo, and &#5196; should have been &#5198;?
If not, let me know in the thank-you note form below, and I'll get back
to you.

     P(A|B) = P(A&#5198;B)/P(B)

       0.55 = P(A&#5198;B)/0.4

(0.55)(0.4) = P(A&#5198;B)

       0.22 = P(A&#5198;B)

That is the answer if you were asked for P(A&#5198;B), not P(A&#5196;B)

-----------------------------------------------------------------
  
FYI, to show that to get P(A&#5196;B) is impossible, let's look at the 
Venn diagram below:

{{{drawing(300,200,-4,4,-2,4.8,rectangle(-4,-1.6,4,4.4), locate(-2,1.8,x),locate(1.5,1.7,z),locate(-3.7,-1,w), locate(-3.6,2.5,A), locate(-.1,1.8,y),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)
 )}}}

Let x,y,z and w be the probabilities of the 4 regions of the Venn
diagram.

P(B)= y+z = 0.4 

P(A|B) = P(AnB)/P(B) = y/(y+z) = y/0.4 = 0.55

 y/0.4 = 0.55
     y = (0.55)(0.4)
     y = 0.22 = P(A&#5198;B)

   y+z = 0.4
0.22+z = 0.4
     z = 0.18 = P(A'&#5198;B)

We know that x+y+z+w = 1
             x+0.22+0.18+w = 1
             x+w = 0.6

{{{drawing(300,200,-4,4,-2,4.8,rectangle(-4,-1.6,4,4.4), locate(-2,1.8,x),locate(1.5,1.7,z=0.18),locate(-3.7,-1,w), locate(-3.6,2.5,A), locate(-.2,2.2,"y="),
locate(-.44,1.6,0.22),

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)
 )}}}

So one possibility for x+w=0.6 is x=0.3 and w=0.3,

we could have:

{{{drawing(300,200,-4,4,-2,4.8,rectangle(-4,-1.6,4,4.4), locate(-2,1.8,x=0.3),locate(1.5,1.7,z=0.18),locate(-3.7,-1,w=0.3), locate(-3.6,2.5,A), locate(-.2,2.2,"y="),
locate(-.44,1.6,0.22),

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)
 )}}} 

Then P(A&#5196;B) = 0.3+0.22+0.18 = 0.7

But another possibility for x+w=0.6 is x=0.4 and w=0.2,

we could have:

{{{drawing(300,200,-4,4,-2,4.8,rectangle(-4,-1.6,4,4.4), locate(-2,1.8,x=0.4),locate(1.5,1.7,z=0.18),locate(-3.7,-1,w=0.2), locate(-3.6,2.5,A), locate(-.2,2.2,"y="),
locate(-.44,1.6,0.22),

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)
 )}}} 

Then P(A&#5196;B) = 0.4+0.22+0.18 = 0.8

And there are also infinitely more possibilities.

Let me know in the thank you note below if you still think you were 
asked to find P(A&#5196;B) and not P(A&#5198;B)
                                     
Edwin</pre>