Very difficult Bayes, smokers problem

Question

50% of people don't smoke, 20% are light smokers & 30% are heavy smokers.

Heavy smokers are twice as likely to die prematurely as light smokers, light smokers are twice as likely to die prematurely as nonsmokers.

What's the probability of being a heavy smoker given person died prematurely?

I'm sure this would be a Bayes' theorem problem, but I can't figure out how to find the probability of deaths.

Answer 1

heavy smokers are twice as likely to die as light smokers, light smokers are twice as likely to die as nonsmokers

that is a succession of frequencies like

$$1;2;4$$

for non smokers, light smokers and heavy smokers that means a probability to die of

$$\left\{\frac{1}{1+2+4};\frac{2}{1+2+4};\frac{4}{1+2+4} \right\}=\left\{\frac{1}{7};\frac{2}{7};\frac{4}{7} \right\}$$

respectively...the rest is a simply Bayes' Theorem example.

Thus, concluding,

$$\mathbb{P}[HS|D]=\frac{0.3\times\frac{4}{7}}{0.5\times\frac{1}{7}+0.2\times\frac{2}{7}+0.3\times\frac{4}{7}}\approx 0.57$$

Answer 2

Let $D$ denote 'die because of some reason such as cancer' as @Joe indicates this is needed since everyone will die with probability $1$. Let $NS$, $L$ and $H$ denote not smoker, light smoker and heavy smoker respectively. Then we have that $P(D|H)=2P(D|L) = 4P(D|NS)$ . We have that $P(NS) = 0.5$, $P(L) = 0.2$ and $P(H)=0.3$.

Then by Bayes' theorem we find that e.g. $P(H|D) = P(D|H)P(H)/P(D)$. Using this and the above we find that by dividing probabilities:

\begin{align} \frac{P(H|D)}{P(L|D)} &= \frac{P(H)}{P(L)} \frac{P(D|H)}{P(D|L)} = (0.3/0.2)\cdot 2 = 3. \end{align}

Similarly, it follows that:

\begin{align} P(L|D) &= 0.8 P(NS|D),\\ P(H|D) &= 1.6 P(NS|D). \\ \end{align}

It holds that $P(L|D) + P(H|D) + P(NS|D) = 1$ (because $L,H,NS$ form a partition of all the people). Hence it follows that $P(H|D)(1+1/3 + 1/(2.4)) = 1 \implies P(H|D) \approx 0.57.$

Answer 3

This question deals strictly with the subspace of the dead. It can be solved quite simply, yet subtly, in this manner:

In the subspace of the dead:

       P(non smoke) + P(light smoke) + P(heavy smoke) = 1

Let us rename thusly:

       P(N) + P(L) + P(H) = 1               (1)

We are supplied the additional linearly independent information that:

       P(L) = 2 * P(N)                      (2)
       P(H) = 2 * P(L)                      (3)

Substitute (2) into (3):

       P(H) = 2 * (2 * P(N))
       P(H) = 4 * P(N)                      (4)

We can now rewrite equation (1) entirely in terms of P(N) using (2) and (4):

       P(N) + (2 * P(N)) + (4 * P(N)) = 1
       7 * P(N) = 1
       P(N) = 1/7                           (5)

Substitute (5) in to (4):

       P(H) = 4 * (1/7)
       P(H) = 4/7

Complete.