Three electric bulbs are chosen at random from 15 bulbs of which 5 are defective. The probability that atleast one is defective is

Options

A. `67/91`
B. `24/91`
C. `57/91`
D. `34/91`

Answer

To find the probability that at least one of the three chosen electric bulbs is defective, we can use the complementary approach. This means we will first calculate the probability that none of the bulbs are defective and then subtract that from 1.

### Step-by-Step Solution:

1. **Identify Total Bulbs and Defective Bulbs:**
- Total bulbs = 15
- Defective bulbs = 5
- Good bulbs = 15 - 5 = 10

2. **Calculate Total Ways to Choose 3 Bulbs:**
- The total number of ways to choose 3 bulbs from 15 is given by the combination formula:
\[ \text{Total outcomes} = \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \]

3. **Calculate Ways to Choose 3 Good Bulbs:**
- The number of ways to choose 3 good bulbs from the 10 good bulbs is:
\[ \text{Favorable outcomes (no defective)} = \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

4. **Calculate Probability of Choosing No Defective Bulbs:**
- The probability that all chosen bulbs are good (none are defective) is:
\[ P(\text{no defective}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{120}{455} \]

5. **Calculate Probability of At Least One Defective Bulb:**
- The probability that at least one bulb is defective is:
\[ P(\text{at least one defective}) = 1 - P(\text{no defective}) = 1 - \frac{120}{455} = \frac{455 - 120}{455} = \frac{335}{455} \]

6. **Simplify the Probability:**
- We can simplify \(\frac{335}{455}\):
- The greatest common divisor (GCD) of 335 and 455 is 65.
- So, \(\frac{335 \div 65}{455 \div 65} = \frac{67}{91}\).

### Final Answer:
The probability that at least one of the three chosen bulbs is defective is:
\[ \frac{67}{91} \]