Smoker Tree Diagram

[dogma]

Hello again!

I'm stuck and need a little push.

Referring to the diagram: out of a group of smokers, 20% are heavy smokers, 30% are light smokers, and 50% are non-smokers. A light smoker is twice as likely to die as a non-smoker but half as likely as a heavy smoker.

H: heavy smoker
L: light smoker
N: non-smoker
D: die

So, I created the tree diagram (see attached) and understand the relationship. I come up with the following.

P(D|L)=2P(D|N)=\frac{1}{2}P(D|H)

P(H|D)=\frac{P(H)P(D|H)}{P(D)}

and likewise for P(L|D) and P(N|D)

I need to figure out what P(H|D) is given the info and the tree I created.

Can someone nudge me in the right direction?

Thanks,
dogma

[dogma]

I think I got it…can someone check me, please?

The probability of a person being a heavy smoker given that he died:

P(H|D)=\frac{P(H \cap D)}{P(D)}=\frac{P(H)P(D|H)}{P(D)}

The probability that a person in this study dies:

P(D)=P(H \cap D)+P(L \cap D)+P(N \cap D)

= P(H)P(D|H)+P(L)P(D|L)+P(N)P(D|N)

…but since P(D|L)= 2P(D|N)= \frac{1}{2}P(D|H)

…so

P(D)= P(H)P(D|H)+\frac{1}{2}P(L)P(D|H)+\frac{1}{4}P(N)P( D|H)

…then

P(H|D)=\frac{P(H)P(D|H)}{ P(H)P(D|H)+\frac{1}{2}P(L)P(D|H)+\frac{1}{4}P(N)P( D|H)}

…the P(D|H)'s cancel out leaving a very simple equation to plug and chug

P(H|D) = 42.1%

[Bartholomew]

Your answer checks.