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 :

Options

A. `7/45`
B. `14/45`
C. `28/45`
D. `8/45`

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a randomly selected person who has a chest disorder is a smoker and non-vegetarian.

### Step-by-Step Solution:

1. **Identify the groups and their sizes**:
- Smokers and non-vegetarians (A): 160
- Smokers and vegetarians (B): 100
- Non-smokers and vegetarians (C): 140

Total number of people = 400.

2. **Calculate the probabilities of each group**:
- Probability of being a smoker and non-vegetarian, \( P(A) = \frac{160}{400} = 0.4 \)
- Probability of being a smoker and vegetarian, \( P(B) = \frac{100}{400} = 0.25 \)
- Probability of being a non-smoker and vegetarian, \( P(C) = \frac{140}{400} = 0.35 \)

3. **Identify the probabilities of having a chest disorder for each group**:
- Probability of having a chest disorder given that the person is a smoker and non-vegetarian, \( P(E|A) = 0.35 \)
- Probability of having a chest disorder given that the person is a smoker and vegetarian, \( P(E|B) = 0.20 \)
- Probability of having a chest disorder given that the person is a non-smoker and vegetarian, \( P(E|C) = 0.10 \)

4. **Calculate the total probability of having a chest disorder, \( P(E) \)**:
\[ P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C) \]
\[ P(E) = (0.35 \times 0.4) + (0.20 \times 0.25) + (0.10 \times 0.35) \]
\[ P(E) = 0.14 + 0.05 + 0.035 = 0.225 \]

5. **Use Bayes' theorem to find the probability that a person is a smoker and non-vegetarian given that they have a chest disorder, \( P(A|E) \)**:
\[ P(A|E) = \frac{P(E|A)P(A)}{P(E)} \]
\[ P(A|E) = \frac{0.35 \times 0.4}{0.225} \]
\[ P(A|E) = \frac{0.14}{0.225} \approx 0.6222 \]

6. **Final answer**:
The probability that the selected person is a smoker and non-vegetarian given that they have a chest disorder is approximately **0.6222** or **62.22%**.