Question 460322
{{{drawing(300,300,-4,4,-3,6,
 locate(-2,1.8,20), locate(1.5,1.7,70),
locate(-3.5,2.5,A), locate(-.4,1.8,50),
red(circle(-sqrt(2),sqrt(2),2)),locate(3.5,2.5,B),
green(circle(sqrt(2),sqrt(2),2))
 )}}} 
<pre>
n(A) 
That's the number of elements in the whole red circle.
There are 20 in the left part of the red circle and 50 
in the right part. So that's 20+50 = 70. 
n(A) = 70 

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n(B)
That's the number of elements in the whole green circle.
There are 50 in the left part of the green circle and 70 
in the right part. So that's 50+70 = 120. 
n(B) = 120 

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

p(A)= n(A)/n(U)

The universal set U consists of all the elements in both
circles which is 20+50+70 or 140.

So p(A) = n(A)/n(U) = 70/140 = 1/2
 
p(B) = n(B)/n(U) = 120/140 = 6/7

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

p(A|B) = The probability of A given B.

Since we are given that we are in B, we can cross out
the 20 because those 20 are not in B, so we just have this:

{{{drawing(300,300,-4,4,-3,6,
 locate(-2,1.8,cross(20)), locate(1.5,1.7,70),
locate(-3.5,2.5,A), locate(-.4,1.8,50),
red(circle(-sqrt(2),sqrt(2),2)),locate(3.5,2.5,B),
green(circle(sqrt(2),sqrt(2),2))
 )}}} 

So p(A|B) = 50/(50+70) = 50/120 = 5/12

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p(B|A) 

Since we are given that we are in A, we can cross out
the 70 because those 70 are not in A, so we just have this:

{{{drawing(300,300,-4,4,-3,6,
 locate(-2,1.8,20), locate(1.5,1.7,cross(70)),
locate(-3.5,2.5,A), locate(-.4,1.8,50),
red(circle(-sqrt(2),sqrt(2),2)),locate(3.5,2.5,B),
green(circle(sqrt(2),sqrt(2),2))
 )}}} 

So p(B|A) = 50/(20+50) = 50/70 = 5/7

Edwin</pre>