Question

A survey of people in a given region showed that $20\%$ were smokers. The probability of death due to lung cancer, given that a person smoked, was $10$ times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is $0.006$, what is the probability of death due to lung cancer given that a person is a smoker?

• A. $\frac{1}{140}$
• B. $\frac{1}{70}$
• C. $\frac{3}{140}$
• D. $\frac{1}{10}$

Answer 1

The correct option is C $\frac{3}{140}$

Let $S$ be the event that person is smoker.
$NS$ be the event that person is non-smoker.
$D$ be the event that death is due to lung cancer.
$P\left(S\right)=\frac{20}{100}=\frac{1}{5}$
$P\left(NS\right)=\frac{4}{5}$
$P\left(D|S\right)=10P\left(D|NS\right)$

Now, using 'Total probability theorem',
$P\left(D\right)=P\left(S\right)\cdot P\left(D|S\right)+P\left(NS\right)\cdot P\left(D|NS\right)$
$\Rightarrow 0.006=\frac{1}{5}\times P\left(D|S\right)+\frac{4}{5}\times P\left(D|NS\right)$
$\Rightarrow \frac{6}{1000}=\frac{1}{5}\times P\left(D|S\right)+\frac{4}{5}\times \frac{1}{10}P\left(D|S\right)$
$\Rightarrow P\left(D|S\right)=\frac{3}{140}$

Therefore, the probability of death due to lung cancer given that a person is a smoker is $\frac{3}{140}$.