**Question:**

In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian.

Their chances of getting a particular chest disorder are $35 \%, 20 \%$ and $10 \%$ respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is :

Correct Option: , 3

**Solution:**

Consider following events

A : Person chosen is a smoker and non vegetarian.

B : Person chosen is a smoker and vegetarian.

$\mathrm{C}$ : Person chosen is a non-smoker and vegetarian.

E : Person chosen has a chest disorder

Given

$P(A)=\frac{160}{400} \quad P(B)=\frac{100}{400} P(C)=\frac{140}{400}$

$P\left(\frac{E}{A}\right)=\frac{35}{100} P\left(\frac{E}{B}\right)=\frac{20}{100} P\left(\frac{E}{C}\right)=\frac{10}{100}$

To find

$P\left(\frac{A}{E}\right)=\frac{P(A) P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)+P(C) \cdot P\left(\frac{E}{C}\right)}$

$=\frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100}+\frac{100}{400} \times \frac{20}{100}+\frac{140}{400} \times \frac{10}{100}}$

$=\frac{28}{45}$ option (3)