In a country, there are 2 states, Jersy and New Castle. 70% of the population are from Jersy, and 30% are from New Castle. Out of the Jersy population, 6% are crazy, and out of New Castle, 3% are crazy. if someone is crazy, what is the probability that he is from Jersy?
So I just said that $P(Jersy|Crazy)= \frac{P(Jersy\cap Crazy}{P(Crazy)}=\frac{0.7*0.06}{0.7*0.06+0.3*0.03} = 0.8235$
Is that correct?
Yes, that is correct.
I would go into more detail than what you have shown, in order to be clear about how you performed the computation.
Let $J$ be the event that a randomly selected person is from Jersy and $N$ be the complementary event that a randomly selected person is from New Castle. Let $C$ be the event that a randomly selected person is crazy.
Then we are asked to determine $$\Pr[J \mid C] = \frac{\Pr[C \mid J]\Pr[J]}{\Pr[C]}.$$ This is Bayes' rule. Then the law of total probability gives for the denominator $$\Pr[C] = \Pr[C \mid J]\Pr[J] + \Pr[C \mid N]\Pr[N].$$ This is exactly what you have computed, with the following probabilities $$\Pr[J] = 0.7, \\ \Pr[C \mid J] = 0.06, \\ \Pr[N] = 1 - \Pr[J] = 0.3, \\ \Pr[C \mid N] = 0.03.$$