Basic question Bayes Theorem, Law of total probability

Question

The question is:

In a given country there are 10% heavy smokers, 20% moderate smokers, and 70% non-smokers. The probability that a randomly chosen inhabitant of this country gets lung cancer is 0.03. Furthermore, moderate smokers have five times higher probability and heavy a ten times higher probability than non-smokers to get lung cancer.

a) Compute the probability that a non-smoker gets lung cancer.

My try,

Let $A :=$ A person gets lungcancer, $B :=$ person is a heavy smoker, $C:=$ person is a moderate smoker $D:=$ person is a non-smoker.

$P(A) = P(A|D)P(D) + P(A|B)P(B) + P(A|C)P(C) $

Because $A,B$ and $C$ partition the sample space.

Now call $P(A|D)=p$ then, $0.03=0.7p+10*0.1p + 5*0.2p$

So $p = 0.03/2.7$

Is this correct?

Answer 1

There was some typo in your definition of your events (you define $C$ as two different events), but the method looks good.