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
$P \left(\frac{ A }{ E }\right)=\frac{ P ( A ) P \left(\frac{ E }{ A }\right)}{ P ( A ) \cdot P \left(\frac{ E }{ A }\right)+ P ( B ) \cdot P \left(\frac{ E }{ B }\right)+ P ( C ) \cdot P \left(\frac{ E }{ C }\right)}$
$=\frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100}+\frac{100}{400} \times \frac{20}{100}+\frac{140}{400} \times \frac{10}{100}}$
$=\frac{28}{45}$