Question

In a group of 400 people, 160 are smokers and non- vegetarian, 100 are smokers and vegetarians and the remaining are non-smokers and vegetarians. The probability of getting a special chest disease are 35%, 20% and 10% respectively. A person chosen from the group at random is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?

Answer 1

Hint – In this question let ${E_1},{E_2}{\text{ and }}{E_3}$ are the special kind of chest disease and $A,B{\text{ and }}C$ are the number of people getting suffered by these diseases respectively. Then find the respective probabilities of getting ${E_1},{E_2}{\text{ and }}{{\text{E}}_3}$ disease respectively. So probability that the randomly selected person having disease which is from the group of smoker and non-vegetarian will be $\dfrac{{{\text{people getting suffered from the disease and from the group of smokers and non - vegetarian}}}}{{{\text{total number of people suffered}}}}$.

Complete step-by-step answer:
Total people = 400
Smokers and non –vegetarian = 160
Smokers and vegetarian = 100
And remaining non-smokers and vegetarians = 400 – 160 – 100 =140.
Now the probabilities of special chest disease are 35%, 20% and 10%.
Let ${E_1},{E_2}{\text{ and }}{E_3}$ are the special kind of chest disease and $A,B{\text{ and }}C$are the number of people getting suffered by these diseases respectively.
Where, ${E_1},{E_2}{\text{ and }}{E_3}$ are the diseases for Smokers and non –vegetarian, Smokers and vegetarian and non-smokers and vegetarian respectively.
Now it is given that ${E_1} = 35$%, ${E_2} = 20$%, and ${E_3} = 10$%
Now as we know that probability is the ratio of favorable number of outcomes to the total number of outcomes,
Therefore probability of getting ${E_1}$ disease = $\dfrac{{35}}{{100}} = \dfrac{A}{{160}}$
So the number of people getting suffered by ${E_1}$ disease = A = 56.
Now, the probability of getting ${E_2}$ disease = $\dfrac{{20}}{{100}} = \dfrac{B}{{100}}$
So the number of people getting suffered by ${E_2}$ disease = B = 20.
And the probability of getting ${E_3}$ disease = $\dfrac{{10}}{{100}} = \dfrac{C}{{140}}$
So the number of people getting suffered by ${E_3}$ disease = C =14.
So the total number of people getting suffered are
$\Rightarrow A + B + C = 56 + 20 + 14 = 90$
Now we have to find out the probability of the randomly selected person having disease which is from the group of smokers and non-vegetarian.
So the probability (P) = $\dfrac{{{\text{people getting suffered from the disease and from the group of smokers and non - vegetarian}}}}{{{\text{total number of people suffered}}}}$
$\Rightarrow \dfrac{A}{{90}} = \dfrac{{56}}{{90}} = \dfrac{{28}}{{45}} = 0.622$
So this is the required probability.

Note – Probability of any event is the ratio of the total favorable outcome for that particular event to that of the total number of possible cases for the entire sample of which this particular event is a part of. The probability of any event always lies between 0 and 1, that is $0 \leqslant P \leqslant 1$. This concept is helpful while solving problems of this kind.