0

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?

RobPratt
  • 45,619
user2323232
  • 672
  • Possibly helpful: https://math.stackexchange.com/questions/2279851/applied-probability-bayes-theorem/2279888#2279888 . Also: reporting four decimal places of precision when input data has only one or two significant figures would make no sense in a real situation, even if correct in this toy problem. – Ethan Bolker Nov 30 '20 at 22:03

1 Answers1

1

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 $$\Pr[J] = 0.7, \\ \Pr[C \mid J] = 0.06, \\ \Pr[N] = 1 - \Pr[J] = 0.3, \\ \Pr[C \mid N] = 0.03.$$

heropup
  • 135,869