A box contains 5 defective and 15 non -defective bulbs . , find the probability that atleast 3 bulbs are defective when 4 bulbs are selected at random .

Options

A. `31/969`
B. `7/20`
C. `1/20`
D. none of these

Answer

To find the probability that at least 3 bulbs are defective when 4 bulbs are selected at random from a box containing 5 defective and 15 non-defective bulbs, we can break the problem down into steps.

### Step-by-Step Solution:

1. **Understand the Total Number of Bulbs:**
- There are 5 defective bulbs and 15 non-defective bulbs.
- Total bulbs = 5 + 15 = 20.

2. **Define the Event:**
- We want to find the probability of selecting at least 3 defective bulbs when 4 bulbs are chosen.
- This can occur in two scenarios:
- Scenario 1: 3 defective bulbs and 1 non-defective bulb.
- Scenario 2: 4 defective bulbs.

3. **Calculate the Total Ways to Choose 4 Bulbs:**
- The total number of ways to choose 4 bulbs from 20 is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \).
- Therefore, the total ways to choose 4 bulbs from 20:
\[ C(20, 4) = \frac{20!}{4!(20-4)!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845. \]

4. **Calculate the Favorable Outcomes for Scenario 1 (3 Defective, 1 Non-defective):**
- The number of ways to choose 3 defective bulbs from 5:
\[ C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10. \]
- The number of ways to choose 1 non-defective bulb from 15:
\[ C(15, 1) = 15. \]
- Therefore, the total ways for Scenario 1:
\[ \text{Total for Scenario 1} = C(5, 3) \times C(15, 1) = 10 \times 15 = 150. \]

5. **Calculate the Favorable Outcomes for Scenario 2 (4 Defective):**
- The number of ways to choose 4 defective bulbs from 5:
\[ C(5, 4) = \frac{5!}{4!(5-4)!} = 5. \]

6. **Total Favorable Outcomes:**
- Total favorable outcomes = Outcomes from Scenario 1 + Outcomes from Scenario 2:
\[ \text{Total Favorable Outcomes} = 150 + 5 = 155. \]

7. **Calculate the Probability:**
- The probability of selecting at least 3 defective bulbs is given by:
\[ P(\text{at least 3 defective}) = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{155}{4845}. \]

8. **Simplify the Probability:**
- To simplify \( \frac{155}{4845} \):
- Both numbers can be divided by 5:
\[ P(\text{at least 3 defective}) = \frac{31}{969}. \]

### Final Answer:
The probability that at least 3 bulbs are defective when 4 bulbs are selected at random is \( \frac{31}{969} \).