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
$A=$the person suffers from the disease
$E_1=$the perso is a smoker and a non-vegetarian
$E_2=$the person is a smoker and a vegetarian
$E_3=$the person is a non-smoker and a vegetarian

$\therefore P\left(E_1\right)=\frac{160}{400}$

$P\left(E_2\right)=\frac{100}{400}$

$P\left(E_3\right)=\frac{140}{400}$

Now,$P\left(A|E_1\right)=\frac{35}{100}$

$P\left(A|E_2\right)=\frac{20}{100}$

$P\left(A|E_3\right)=\frac{10}{100}$

Using Bayes' theorem, we get

Required probability$=P\left(E_1|A\right)=\frac{P\left(E_1\right)P\left(A|E_1\right)}{P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)+P\left(E_3\right)P\left(A|E_3\right)}$

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

$=\frac{560}{560+200+140}=\frac{560}{900}=\frac{28}{45}$