Question

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:

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

Hint:

Bayes' theorem problems are often easier to solve by first calculating the number of people in each sub-category.
Number with disorder from A: 160 0.35 = 56.
Number with disorder from B: 100 0.20 = 20.
Number with disorder from C: 140 0.10 = 14.
Total with disorder = 56+20+14 = 90.
Prob(from A | has disorder) = (Number from A with disorder) / (Total with disorder) = 56/90 = 28/45.

Answer 1

The Correct Option is C

Let A: Smoker & Non-vegetarian B: Smoker & Vegetarian C: Non-smoker & Vegetarian D: Chest disorder \[ P(A)=\frac{160}{400},\; P(B)=\frac{100}{400},\; P(C)=\frac{140}{400} \] \[ P(D|A)=0.35,\; P(D|B)=0.20,\; P(D|C)=0.10 \] Step 1: Total probability \[ P(D)=0.35\cdot\frac{160}{400} +0.20\cdot\frac{100}{400} +0.10\cdot\frac{140}{400} =\frac{9}{40} \] Step 2: Apply Bayes’ theorem \[ P(A|D)=\frac{P(D|A)P(A)}{P(D)} = \frac{0.35\cdot\frac{160}{400}}{\frac{9}{40}} =\frac{28}{45} \] \[ \boxed{\frac{28}{45}} \]