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 :
Select the correct option:
A
$\frac{7}{45}$
B
$\frac{14}{45}$
C
$\frac{28}{45}$
D
$\frac{8}{45}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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
$$
\begin{aligned}
& P(A)=\frac{160}{400} P(B)=\frac{100}{400} P(C)=\frac{140}{400} \\
& P\left(\frac{E}{A}\right)=\frac{35}{100} P\left(\frac{E}{B}\right)=\frac{20}{100} P\left(\frac{E}{C}\right)=\frac{10}{100}
\end{aligned}
$$
To find
$$
\begin{aligned}
& P\left(\frac{A}{E}\right)=\frac{P(A) P\left(\frac{E}{A}\right)}{P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)+P(C) \cdot P\left(\frac{E}{C}\right)} \\
& =\frac{\frac{160}{400} \times \frac{35}{100}}{\frac{160}{400} \times \frac{35}{100}+\frac{100}{400} \times \frac{20}{100}+\frac{140}{400} \times \frac{10}{100}} \\
& =\frac{28}{45}
\end{aligned}
$$
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