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?
Given:
From 400 people
Smokers and non - vegetarian are 160
Smokers and vegetarian are 100
Non - smokers and vegetarian are 140
Let us assume U1, U2, U3 and A be the events as follows:
U1 = choosing Smokers and non - vegetarian
U2 = choosing Smokers and vegetarian
U3 = choosing Non - smokers and vegetarian
A = getting special chest disease
From the problem
⇒ 
⇒ 
⇒ 
⇒ P(A|U1) = P(Smoker and non - vegetarian getting Chest disease)
⇒ 
⇒ P(A|U2) = P(Smoker and vegetarian getting chest disease)
⇒ 
⇒ P(A|U3) = P(Non - smoker and vegetarian getting chest disease)
⇒ 
Now we find
P(U1|A) = P(The selected chest diseased person is a smoker and non - vegetarian)
Using Baye’s theorem:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
∴ The required probability is 
.