In a batch of 10 light bulbs , 2 are defective . If 3 of the bulbs are chosen at random, what is the probability that at least 1 of the chosen bulbs is defective ?

Options

A. `8/15`
B. `7/15`
C. `3/10`
D. `1/4`

Answer

To solve the problem of finding the probability that at least one of the chosen bulbs is defective when selecting 3 bulbs from a batch of 10 (where 2 are defective), we can follow these steps:

### Step 1: Understand the Total and Defective Bulbs
We have a total of 10 light bulbs, out of which 2 are defective and 8 are non-defective.

### Step 2: Define the Event
We want to find the probability that at least one of the chosen bulbs is defective. This can be calculated using the complement rule:
\[ P(\text{at least 1 defective}) = 1 - P(\text{no defective}) \]

### Step 3: Calculate the Probability of Choosing No Defective Bulbs
To find \(P(\text{no defective})\), we need to calculate the probability of choosing 3 non-defective bulbs from the 8 available non-defective bulbs.

The number of ways to choose 3 non-defective bulbs from 8 is given by the combination formula:
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

### Step 4: Calculate the Total Ways to Choose 3 Bulbs from 10
The total number of ways to choose any 3 bulbs from the 10 available is:
\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \]

### Step 5: Calculate the Probability of Choosing No Defective Bulbs
Now, we can find the probability of choosing no defective bulbs:
\[ P(\text{no defective}) = \frac{\text{Number of ways to choose 3 non-defective}}{\text{Total ways to choose 3 bulbs}} = \frac{56}{120} \]

### Step 6: Simplify the Probability
We can simplify \( \frac{56}{120} \):
\[ \frac{56}{120} = \frac{14}{30} = \frac{7}{15} \]

### Step 7: Calculate the Probability of At Least One Defective Bulb
Now, we can find the probability of at least one defective bulb:
\[ P(\text{at least 1 defective}) = 1 - P(\text{no defective}) = 1 - \frac{7}{15} = \frac{8}{15} \]

### Conclusion
Thus, the probability that at least one of the chosen bulbs is defective is:
\[ \boxed{\frac{8}{15}} \]