3 bulbs are chosen at random from 20 bulbs , out of which 4 are defective . The probability that the room is illuminated will be

Options

A. `(282)/(283)`
B. `(283)/(285)`
C. `(284)/(285)`
D. `(281)/(285)`

Answer

To solve the problem of finding the probability that the room is illuminated when 3 bulbs are chosen at random from a total of 20 bulbs (of which 4 are defective), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Total Bulbs and Defective Bulbs**:
- Total bulbs = 20
- Defective bulbs = 4
- Working bulbs = 20 - 4 = 16

2. **Understand the Condition for Illumination**:
- The room will be illuminated if at least one of the chosen bulbs is working.

3. **Calculate the Probability of Choosing Only Defective Bulbs**:
- We need to find the probability of choosing 3 defective bulbs (which means the room will not be illuminated).
- The number of ways to choose 3 defective bulbs from 4 is given by the combination formula \( \binom{n}{r} \).
- Thus, the number of ways to choose 3 defective bulbs from 4 is \( \binom{4}{3} \).

4. **Calculate the Total Ways to Choose Any 3 Bulbs**:
- The total number of ways to choose any 3 bulbs from 20 is \( \binom{20}{3} \).

5. **Calculate the Probability of Choosing 3 Defective Bulbs**:
- The probability of choosing 3 defective bulbs is given by:
\[ P(\text{3 defective}) = \frac{\binom{4}{3}}{\binom{20}{3}} \]

6. **Calculate the Complement Probability**:
- To find the probability that at least one bulb is working, we subtract the probability of choosing 3 defective bulbs from 1:
\[ P(\text{at least 1 working}) = 1 - P(\text{3 defective}) \]

7. **Substituting Values**:
- Calculate \( \binom{4}{3} = 4 \) and \( \binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140 \).
- Therefore:
\[ P(\text{3 defective}) = \frac{4}{1140} \]

8. **Final Calculation**:
- Now calculate:
\[ P(\text{at least 1 working}) = 1 - \frac{4}{1140} = \frac{1140 - 4}{1140} = \frac{1136}{1140} \]
- Simplifying \( \frac{1136}{1140} \) gives:
\[ P(\text{at least 1 working}) = \frac{284}{285} \]

### Final Answer:
The probability that the room is illuminated is \( \frac{284}{285} \).