Question 1160377
<pre><font size=4><b>
{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,3,""*3),
locate(-2.3,2,""*1),
locate(3,-2.5,S),

locate(0,-2.7,1),
locate(-.45,-1,""*4),
locate(.6,.4,""*2), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,C),

locate(-.4,2.3,""*5),
locate(1.8,2,""*6) )}}}

I've replaced the points with their probabilities"

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,3,"0.1"),
locate(-2.3,2,0.3),
locate(3,-2.5,S),

locate(0,-2.7,B),
locate(-.45,-1,"0.1"),
locate(.6,.4,"0.2"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,C),

locate(-.4,2.3,"0.1"),
locate(1.8,2,"0.2") )}}}

a) P(A) = sum of probabilities in circle A = 0.1+0.3+0.1 = 0.5

b) P(BᑎA) = sum of probabilities in the overlapping part of 
circle B and A.  This is totally empty so the probability = 0.

c) P(AᑌBᑌC) = sum of all probabilities in circles A, B and C =
0.1+0.3+0.1+0.2+0.2+0.1 = 1.0

d) P(C') = sum of all probabilities EXCEPT those in circle C = 
0.1+0.3+.1 = 0.5

e) P(AᑎC') = sum of probabilities in A which are not in circle C. 
0.1+0.3=0.4

f) P(B|A) = P(BᑎA)/P(A)

P(BᑎA) = 0, from b) above
P(A) = 0.5, from a) above

So P(B|A) = P(BnA)/P(A) = 0/0.5 = 0

f) Are A and C mutually exclusive?

No because the overlapping part of A and C is not empty.

Edwin</font></b></pre>