Question

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 special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?

Answer 1

Let A, E₁, E₂ and E₃ denote the events that the person suffers from the disease, is a smoker and a non-vegetarian, is a smoker and a vegetarian and the person is a non-smoker and a vegetarian, respectively.

\[\therefore P\left( E_1 \right) = \frac{160}{400}\]
\[ P\left( E_2 \right) = \frac{100}{400} \]
\[ P\left( E_3 \right) = \frac{140}{400}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{35}{100}\]
\[P\left( A/ E_2 \right) = \frac{20}{100}\]
\[P\left( A/ E_3 \right) = \frac{10}{100}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100} + \frac{100}{400} \times \frac{20}{100} + \frac{140}{400} \times \frac{10}{100}}\]
\[ = \frac{560}{560 + 200 + 140} = \frac{560}{900} = \frac{28}{45}\]