The chance of defective screws in three boxes `A ,B ,Ca r e1//5,1//6,1//7,` respectively. A box is selected at random and a screw draw in from it at random is found to be defective. Then find the probability that it came from box `Adot`

Explanation

To solve the problem, we will use Bayes' theorem. We need to find the probability that a defective screw came from box A given that a defective screw was drawn.

### Step-by-step Solution:

1. **Define Events**:
- Let \( E_1 \): Event that box A is selected.
- Let \( E_2 \): Event that box B is selected.
- Let \( E_3 \): Event that box C is selected.
- Let \( A \): Event that a screw drawn is defective.

2. **Probabilities of Selecting Each Box**:
Since a box is selected at random, the probabilities of selecting each box are:
\[ P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \]

3. **Probabilities of Drawing a Defective Screw**:
- Probability of drawing a defective screw from box A:
\[ P(A | E_1) = \frac{1}{5} \]
- Probability of drawing a defective screw from box B:
\[ P(A | E_2) = \frac{1}{6} \]
- Probability of drawing a defective screw from box C:
\[ P(A | E_3) = \frac{1}{7} \]

4. **Total Probability of Drawing a Defective Screw**:
We need to find \( P(A) \):
\[ P(A) = P(E_1) \cdot P(A | E_1) + P(E_2) \cdot P(A | E_2) + P(E_3) \cdot P(A | E_3) \]
Substituting the values:
\[ P(A) = \frac{1}{3} \cdot \frac{1}{5} + \frac{1}{3} \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{1}{7} \]
\[ P(A) = \frac{1}{15} + \frac{1}{18} + \frac{1}{21} \]

To add these fractions, we find a common denominator, which is 630:
\[ P(A) = \frac{42}{630} + \frac{35}{630} + \frac{30}{630} = \frac{107}{630} \]

5. **Using Bayes' Theorem**:
We want to find \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(E_1) \cdot P(A | E_1)}{P(A)} \]
Substituting the values:
\[ P(E_1 | A) = \frac{\frac{1}{3} \cdot \frac{1}{5}}{\frac{107}{630}} \]
\[ P(E_1 | A) = \frac{\frac{1}{15}}{\frac{107}{630}} = \frac{630}{15 \cdot 107} = \frac{42}{107} \]

### Final Answer:
The probability that the defective screw came from box A is:
\[ \frac{42}{107} \]