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A health study tracked a group of persons for five years. At the beginning of the study, 20% were classified as heavy smokers, 30% as light smokers, and 50% as non-smokers. Results of the study showed that light smokers were twice as likely as non-smokers to die during the five-year study, but only half as likely as heavy smokers. Calculate the probability that a participant was a heavy smoker if the participant died during the five-year study.

I tried this using Bayes' theorem

We need to find $P(H|D)$ where $H$ is the event that the participant is a heavy smoker, $N$ a non-smoker, and $L$ a light smoker.

It is given that $P(H)=0.2,P(L)=0.3,P(N)=0.5$

$$ P(D|L)=2P(D|N)=0.5P(D|H), (1)\\ P(D|L)+P(D|N)+P(D|H) = 1 $$

$$ P(H|D)=\dfrac{P(D|H)P(H)}{P(D|N)P(N)+P(D|L)P(L)+P(D|H)P(H)}, (2) $$

Substituting equation $(1)$ in equation $(2)$ simplifying and plugging in values I got $P(H|D)=0.26$, whereas the answer in my textbook is 0.42.

Where did I go wrong?

ananta
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Noro
  • 11

2 Answers2

1

The correct answer follows:

Let $H \equiv $ the person in case is a heavy smoker, $L \equiv $ the person in case is a lightsmoker and $N \equiv $ the person in case is a nonsmoker. Also, let $D \equiv $ the person in case died over the five year period.

From the information given in the exercise, you know that $$ P(H) = 0.20, \, P(L) = 0.30, \, P(N) = 0.50$$ and also that $P(D \,|\, L) = 2 P(D \, | \, N) = 0.50P(D \, | \, H).$ This obviously implies that $P(D \, | \, N) = 0.25P(D \, |\, H)$. To solve the exercise, you are looking to compute $P( H \, | \, D).$ Using Bayes formula,

\begin{equation*} \begin{split} P(H \, | \, D) = \frac{P(HD)}{P(D)} &= \frac{P(D \, | \, H) P(H)}{P(D \, | \, N)P(N) + P(D \, | \, L)P(L) + P(D \, | \, H) P(H) } \\[.5cm] &= \frac{P(D \, | \, H) 0.20}{0.125P( D \, | \, H) + 0.15P(D \, | \, H) + 0.20 P(D \, | \, H)} \\[.5cm] &= \frac{0.20}{0.125+0.15+0.20} \approx 0.42. \end{split} \end{equation*}

If you have any further doubt about my answer let me know.

xyz
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Your method was perfect. You probably simply made a mistake in your calculations, which you don't show.

Call $a$ the number $P(D|L)=2P(D|N)=0.5P(D|H)$ of your equation 1.

Substituting in your equation 2 gives

$$P(H|D)=\frac{2a\cdot0.2}{0.5a\cdot0.5+a\cdot0.3+2a\cdot0.2}=\frac{0.4a}{0.95a}\approx0.42$$

Anne Bauval
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  • In reality, it was not. You edited the original post to make it look like the method was perfect, wrongly. Initially, the OP assumed that he was given $$ P(H)=0.2P(L)=0.3P(N)=0.5 , $$ And this is not the same as your edition of the question. – xyz Dec 13 '22 at 14:52
  • I just added the two obviously missing commas. This was a minor and irrelevant formatting problem. – Anne Bauval Dec 13 '22 at 14:54
  • I understand your point but "obviously missing" is arguable... at least. – xyz Dec 13 '22 at 14:55
  • I don't need the least to "argue". Let us see what the OP says ;-) – Anne Bauval Dec 13 '22 at 14:56