25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer then a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $$\frac{k}{10}$$. Then the value of $$k$$ is _____.

Let S = smoker, NS = non-smoker, C = lung cancer.

P(S) = 0.25, P(NS) = 0.75

P(C|S) = 27 × P(C|NS). Let P(C|NS) = p, so P(C|S) = 27p.

Using Bayes' theorem:

$$P(S|C) = \frac{P(C|S) \times P(S)}{P(C|S) \times P(S) + P(C|NS) \times P(NS)}$$

$$= \frac{27p \times 0.25}{27p \times 0.25 + p \times 0.75}$$

$$= \frac{6.75p}{6.75p + 0.75p} = \frac{6.75}{7.5} = \frac{27}{30} = \frac{9}{10}$$

So $$P(S|C) = \frac{9}{10} = \frac{k}{10}$$

Therefore $$k = 9$$.