Conditional probability using venn diagram? I'm a bit confused about the conditional probability. I tried to prove using venn diagrams that if $P(A) = P(B|A)$, then $P(B) = P(A)$. Why is this not true?
For example, I have some conditional probability like this: $P(A) = \frac15$ and $P(B|A) = \frac15$, isn't $P(B) = \frac15?$ 
If you draw a venn diagram, you will see that B must be equal to A because $P(B|A)$ is equivalent to $P(A)$. In other words B, exactly overlaps with A, right?
 A: The standard formula is  
$$P(B|A)=\frac {P (A \cap B) }{P (A) }$$
Thus, $P (A)=P(B | A) \,\,\,$ implies $$P (A \cap B) =[P (A)]^2\,\,\,$$ and not $P (A)=P (B)\,\,$.
For example, consider a Venn diagram where there is one circle representing $P(A) =0.2\,\,\,$, a second circle representing $P (B) =0.9\,\,\,$, and where the two circles overlap in a relatively small region whose area is equal to $0.04\,$. In this case, we have
$$P(B|A)=\frac {P (A \cap B) }{P (A) }=\frac{0.04}{0.2}=0.2$$
and then $P (B|A)\, $ is equal to $P (A) \,$, but clearly we have  $P (A) \neq P (B) \,\,\,$.
In other words, the problem in your considerations is when you conclude that, if $P (B|A) =P (A)\,\, $, then $B $ must exactly overlap $A$. In fact, the condition that $P (B|A)\,\,$ is equal to $P (A) $ only tells us that the ratio between the area of the intersection of the two circles and the area of the circle representing $A $ is equal to the numerical value of the circle A. It does not tell us that the intersection area or the area of circle $B $ have to be exactly overlapped with that of circle $A $.
A: On a Venn Diagram, $P(B/A)$ can be pictured as selecting from the circle representing $B$ given that your selection is also from $A.$ Hence, you are selecting an element of $A \cap B$ in the context where your universe of possible outcomes is restricted to the circle $A$ (which is how to picture conditional probability on a Venn diagram). In this case, $P(B/A)=P(A)$ means that selecting $B$ given that you're selecting from $A$ is the same probability of selecting from $A$ in the parent Venn diagram; thus, $A \cap B$ consists of the same proportion of $A$ as $A$ does to the original sample space. So long as $A$ itself wasn't equal to the original sample space, then $B \cap A \not = A.$ 
Also, consider cases where $B-A$ is nonempty. Then there is a portion of the circle representing $B$ that does NOT overlap $A$ on the Venn diagram. There are many counterexamples you can construct using this information to overturn your initial intuition. 
A: iF A bs B are independent then  then  P(A∩B) = p(A)p(B). Therefore in case of independence of A and B, if  P(A)=P(B|A) then P(A)=P(B|A)=P(A∩B)/P(A) = P(B).Otherwise it does not have to be true. 
