struggling in visualizing the set relation to calculate probability Given this problem, "Suppose that 55% of all adults regularly consume coffee, 45% regularly consume carbonated soda, and 70% regularly consume at least one of the two. "  I have two questions to solve. The first, "What is the probability that a randomly selected adult regularly consumes both?"  So,   let $A$ be adults consuming coffee and $B $ be those consuming carbonated soda. I know that I'm being asked to calculate $A \cap B$ but I'm not quite seeing how to visualize it into the pattern given in the book, $P(A \cup B)=P (A) +P (B)-P (A \cap B) $.  I've got the following which I hope means I'm on the right track $A= A\cap B + A \cap B'$ and $B=A\cap B+A'\cap B $.  Am I, or should I be looking somewhere else for alternate ways of representation to calculate the probability of $A\cap B $?
Thanks
 A: Take a look at the following Venn Diagram:

Note that:
$A = I \cup II$
$B = II \cup III$
$A \cap B = II$
$A \cup B = I \cup II \cup III$
Now, if you count everything that is inside the $A$ circle ($I$ + $II$), and you add to this everything that is inside the $B$ circle ($II$ + $III$), then you have counted everything that is inside $A \cap B$ ($II$) twice, so to count everything that is inside the union of $A$ and $B$ ($I + II + III$), you need to subtract everything that is inside $A \cap B$ ($II$) once. Hence:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 
From this, it follows that:
$$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.55 + 0.45 - 0.7 = 0.3$$ 
A: Comment:
Looking at a Venn Diagram is often helpful, and you may learn something 
about this particular problem by
assigning a probability to each of the regions of a properly drawn
Venn Diagram [which would also have a region for $(A\cap B)^\prime$, 
'regularly consumes neither type of beverage'].
I believe this problem was supposed to reinforce the 'general addition rule'
$P(A\cup B) = P(A) + P(B) - P(A\cap B),$ as follows:
$.70 = P(A\cup B) = P(A) + P(B) - P(A\cap B) = .55 + .45 - P(A\cap B)
= 1 - P(A\cap B),$
which implies $P(A\cap B) = .30.$
Note: Also, $P[(A \cup B)^\prime] = P(A^\prime \cap B^\prime) = .30.$
