In a box, there are 12 defective light bulbs and 18 non-defective light bulbs. If you pick 4 light bulbs at random without replacement, what is the probability that exactly 2 of them are defective?
Let's break down this problem step-by-step:
1. We know there are 12 defective light bulbs and 18 non-defective light bulbs in the box, for a total of 30 light bulbs.
2. We want to find the probability of picking 4 light bulbs without replacement such that exactly 2 of them are defective.
3. There are C(12, 2) ways to choose 2 defective light bulbs from the 12 defective ones, which equals 66 combinations.
4. Similarly, there are C(18, 2) ways to choose 2 non-defective light bulbs from the 18 non-defective ones, which equals 153 combinations.
5. Therefore, there are 66 * 153 = 10,098 ways to pick 4 light bulbs with exactly 2 of them being defective.
6. The total number of ways to pick any 4 light bulbs from the 30 light bulbs is C(30, 4), which equals 27,405 combinations.
7. Finally, the probability of picking 4 light bulbs without replacement and having exactly 2 of them be defective is the number of ways to do so divided by the total number of ways to pick any 4 light bulbs: 10,098 / 27,405 ≈ 0.368.

The probability that exactly 2 of the 4 randomly picked light bulbs are defective is approximately 0.368.