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 :

• A. $\frac{7}{45}$
• B. $\frac{14}{45}$
• C. $\frac{28}{45}$
• D. $\frac{8}{45}$

Answer 1

Answer: (C)

Consider following events
A : Person chosen is a smoker and non vegetarian.
$B$ : Person chosen is a smoker and vegetarian.
C : Person chosen is a non-smoker and vegetarian.
$E:$ Person chosen has a chest disorder
Given
$P(A)=\frac{160}{400}P(B)=\frac{100}{400}P(C)=\frac{140}{400}$
$P\left(\frac{E}{A}\right)=\frac{35}{100}P\left(\frac{E}{B}\right)=\frac{20}{100}P\left(\frac{E}{C}\right)=\frac{10}{100}$
To find
$P\left(\frac{A}{E}\right)=\frac{P(A)P\left(\frac{E}{A}\right)}{P(A)\cdot P\left(\frac{E}{A}\right)+P(B)\cdot P\left(\frac{E}{B}\right)+P(C)\cdot P\left(\frac{E}{C}\right)}$
$=\frac{\frac{160}{400}\times \frac{35}{100}}{\frac{160}{400}\times \frac{35}{100}+\frac{100}{400}\times \frac{20}{100}+\frac{140}{400}\times \frac{10}{100}}$
$=\frac{28}{45}$