A lot of 60 bulbs contain 12 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Suppose the bulb drawn in first attempt is defective and is not replaced. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Answer

To solve the problem step by step, we will break it down into two parts as described in the question.

### Part 1: Probability that the bulb drawn is defective

1. **Identify the total number of bulbs and defective bulbs:**
- Total bulbs = 60
- Defective bulbs = 12

2. **Calculate the probability of drawing a defective bulb:**
- Probability (P) of drawing a defective bulb is given by the formula:
\[ P(\text{defective}) = \frac{\text{Number of defective bulbs}}{\text{Total number of bulbs}} = \frac{12}{60} \]

3. **Simplify the fraction:**
- Simplifying \(\frac{12}{60}\):
\[ \frac{12 \div 12}{60 \div 12} = \frac{1}{5} \]

Thus, the probability that the bulb drawn is defective is \(\frac{1}{5}\).

### Part 2: Probability that the bulb drawn is not defective after one defective bulb is removed

1. **Update the total number of bulbs and defective bulbs after one defective bulb is drawn:**
- After drawing one defective bulb, the total number of bulbs left = 60 - 1 = 59
- Defective bulbs left = 12 - 1 = 11
- Non-defective bulbs = Total bulbs - Defective bulbs = 59 - 11 = 48

2. **Calculate the probability of drawing a non-defective bulb:**
- Probability (P) of drawing a non-defective bulb is given by the formula:
\[ P(\text{not defective}) = \frac{\text{Number of non-defective bulbs}}{\text{Total number of bulbs left}} = \frac{48}{59} \]

Thus, the probability that the bulb drawn is not defective after one defective bulb has been removed is \(\frac{48}{59}\).

### Summary of Results:
- Probability that the bulb drawn is defective: \(\frac{1}{5}\)
- Probability that the bulb drawn is not defective after one defective bulb is removed: \(\frac{48}{59}\)