Basic Doubt in conditional probability 
Let there are 2 urns. In Urn 1, there are 5 blue balls and 7 red balls. In Urn 2, there are 3 blue balls and 4 red balls. A: Urn 1 is chosen. B is ball is red. What is P(A/B) and P(B/A)

My approach:
P(B/A) = In urn 1 what is probability of picking a red ball. = 7/12
If i apply conditional probability, $P(B/A)= \frac{P(B \cap A)}{P(A)} \\ 
 = \frac{\text{In urn1, red ball probability}}{\text{probability of picking urn 1}} \\=\frac{7/12}{1/2} \\ = \frac{7}{6}$
But answer is given as 7/12. I am not able to understand mistake i did in method using conditional probability.
P(A/B) --> given red ball, what is probability that it has come from urn 1
P(B) = probability of red ball = either it has come from urn 1 or from urn 2 = (1/2)(7/12)+(1/2)(4/7) = 97/168
$ P(A/B)= \frac{P(B \cap A)}{P(B)} \\ 
 = \frac{\text{probability of urn 1's red ball}}{\text{probability of picking red ball}} \\ = \frac{7/12}{97/168} \\ = \frac{98}{97}$ 
Again nonsense answer. 
In text book he applied conditional probability - $P(A/B ) =  \frac{P(B / A)*P(A)}{P(B)}$ --> i dont know from where to get the intuition that i need to apply this. There is also another conditional probability formula in terms of intersection(as in begining of my aprroach. I understood the derivation)
Please point where i did mistake. I gave a thought but could not understand. I am not good in probability area.
 A: $P(B \cap A)$ is the probability that you choose pick a red ball from the first urn, which is the probability you pick the first urn and then a red ball, which is $\frac12 \times \frac7{12}=\frac7{24}$
Your $\frac7{12}$ is the probability you pick a red ball given that you picked the first urn, which is $P(B \mid A)$
More generally $P(A)P(B \mid A) = P(B \cap A) = P(B)P(A \mid B)$ and division will give you the results you need, providing that none of the probabilities are zero, perhaps also using $P(B)= P(B \cap A)+ P(B \cap A^c) = P(A)P(B \mid A)+P(A^c)P(B \mid A^c)$ 
A: P(B|A) = P(A and B) / P(A)
$\frac{1}{2}$ chance of selecting urn 1, then $\frac{7}{12}$ chance selecting red from the urn
P(A and B) = $\frac{1}{2} \times \frac{7}{12} = \frac{7}{24}$
$P(A) = \frac{1}{2}$
P(B|A) = P(A and B) / P(A)
= $\frac{7}{24} / \frac{1}{2} = \frac{7}{12}$
A: You have a bit of an error in your method here. The intersection $P(A\cap B)$ is not 7/12, that would be only the probability that $B$ given $A$, but this misses the probability of being in urn 1. (Presumably you used 1/2 and this would balance the ratio)
Similarly you have an error in your $P(B|A)$...
As for intuition:
$$P(A \cap B) = P(A|B)P(B)$$
$$P(A \cap B) = P(B|A)P(A)$$
Minor Algebra:
$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$
