Question

In a group of $400$ people, $160$ are smokers and non-vegetarians; $100$ are smokers and vegetarians and the remaining $140$ are non-smokers and vegetarians. Their chances of getting a particular chest disorder are $35\%,20\%\mathrm{and}10\%$ respectively. A person is chosen from the group at random and is found to be suffering from chest disorder. The probability that the selected person is a smoker and non-vegetarian is:

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

Answer 1

The correct option is D

$\begin{array}{l}\frac{28}{45}\end{array}$

Explanation for the correct answer:

Find the probability.

Given,

$160$ are smokers as well as non-vegetarians.

$100$ are smokers as well as vegetarians.

$140$ are non-smokers as well as vegetarians.

Their possibilities of getting a particular chest disorder are $35\%,20\%\mathrm{and}10\%$ .

Thus,

The number of smokers, as well as a non-vegetarians suffering from chest disorder, are $n\left(E_1\right)=160\times \frac{35}{100}=56$

The number of smokers, as well as vegetarians suffering from chest disorder, are $n\left(E_2\right)=100\times \frac{20}{100}=20$

The number of non-smokers, as well as vegetarians suffering from chest disorder, are $n\left(E_3\right)=140\times \frac{10}{100}=14$

Therefore the total number of people suffering from the chest disorder are

$\begin{array}{rcl}n\left(S\right)& =& n\left(E_1\right)+n\left(E_2\right)+n\left(E_3\right)\\ & =& 56+20+14\\ & =& 90\end{array}$

The required probability that the opted person is a smoker as well as a non-vegetarian is,

$\begin{array}{rcl}p\left(E_1\right)& =& \frac{n\left(E_1\right)}{n\left(S_1\right)}\\ & =& \frac{56}{90}\\ & =& \frac{28}{45}\end{array}$

Hence, the correct option is D.