In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian, and the remaining are non-smokers and vegetarian. The probabilities of getting a certain chest disease for these three kinds of people are 35%, 20% and 10% respectively. A person is chosen at random from the group and is found to be suffering from the disease. The probability that the person is a smoker is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{38}{45}\)
Let \(E_1\), \(E_2\), \(E_3\) be the events that the person is a smoker and non-vegetarian, a smoker and vegetarian, and a non-smoker and vegetarian, and let D be the event that the person has the disease. There are \(400-160-100 = 140\) people of the third kind.
\( \displaystyle P(E_1) = \frac{160}{400},\quad P(E_2) = \frac{100}{400},\quad P(E_3) = \frac{140}{400} \)
\(P(D|E_1) = 0.35,\quad P(D|E_2) = 0.20,\quad P(D|E_3) = 0.10\)
The common factor \(\frac{1}{400}\) cancels in Bayes’ theorem, so the products can be taken as numbers of people with the disease:
\(160\times 0.35 = 56,\quad 100\times 0.20 = 20,\quad 140\times 0.10 = 14\)
\(56+20+14 = 90\)
A smoker is a person of the first or the second kind, so the two posterior probabilities are added:
\( \displaystyle P(E_1|D)+P(E_2|D) = \frac{56+20}{90} = \frac{76}{90} = \frac{38}{45} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{28}{45}\) — Incorrect. This is \(\frac{56}{90} = P(E_1|D)\), smokers who are non-vegetarian only.
Option (2) \(\displaystyle\frac{13}{20}\) — Incorrect. This is \(\frac{260}{400}\), the share of smokers in the whole group, before the disease is known.
Option (4) \(\displaystyle\frac{7}{45}\) — Incorrect. This is \(\frac{14}{90} = P(E_3|D)\), the probability that the person is a non-smoker.