In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. On a given day, one of the three persons A, B and C carries out this task. A has 45% chance, B has 35% chance and C has 20% chance of doing the task. The probability that A, B and C will take more than the allotted time is `(1)/(6), (1)/(10) and (1)/(20)` respectively. If it is found that the time taken is more than the allotted time, what is the probability that A has done the task ?

Answer

To solve the problem, we will use Bayes' theorem. We need to find the probability that person A has done the task given that the time taken is more than the allotted time.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( E_1 \) be the event that person A does the task.
- Let \( E_2 \) be the event that person B does the task.
- Let \( E_3 \) be the event that person C does the task.
- Let \( H \) be the event that the time taken is more than the allotted time.

2. **Given Probabilities**:
- \( P(E_1) = 0.45 \) (Probability that A does the task)
- \( P(E_2) = 0.35 \) (Probability that B does the task)
- \( P(E_3) = 0.20 \) (Probability that C does the task)
- \( P(H|E_1) = \frac{1}{6} \) (Probability that time is more than allotted time given A does the task)
- \( P(H|E_2) = \frac{1}{10} \) (Probability that time is more than allotted time given B does the task)
- \( P(H|E_3) = \frac{1}{20} \) (Probability that time is more than allotted time given C does the task)

3. **Apply Bayes' Theorem**:
We want to find \( P(E_1|H) \), which is given by Bayes' theorem:
\[ P(E_1|H) = \frac{P(H|E_1) \cdot P(E_1)}{P(H)} \]
where \( P(H) \) can be calculated using the law of total probability:
\[ P(H) = P(H|E_1) \cdot P(E_1) + P(H|E_2) \cdot P(E_2) + P(H|E_3) \cdot P(E_3) \]

4. **Calculate \( P(H) \)**:
\[ P(H) = \left(\frac{1}{6} \cdot 0.45\right) + \left(\frac{1}{10} \cdot 0.35\right) + \left(\frac{1}{20} \cdot 0.20\right) \]
\[ = \frac{0.45}{6} + \frac{0.35}{10} + \frac{0.20}{20} \]
\[ = 0.075 + 0.035 + 0.01 = 0.120 \]

5. **Substitute Values into Bayes' Theorem**:
\[ P(E_1|H) = \frac{P(H|E_1) \cdot P(E_1)}{P(H)} = \frac{\left(\frac{1}{6} \cdot 0.45\right)}{0.120} \]
\[ = \frac{0.075}{0.120} = \frac{75}{120} = \frac{5}{8} \]

6. **Final Answer**:
The probability that A has done the task given that the time taken is more than the allotted time is \( \frac{5}{8} \).