From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

To find the probability distribution of the number of defective bulbs, we can use the concept of binomial distribution.

In this case, we have a lot of 30 bulbs, including 6 defectives. We are drawing a sample of 4 bulbs at random with replacement, which means that after each draw, the bulb is placed back into the lot before the next draw.

The probability of drawing a defective bulb in each individual draw is given by the ratio of the number of defective bulbs (6) to the total number of bulbs (30), which is 6/30 = 1/5.

Now, let's calculate the probability distribution for the number of defective bulbs in the sample:

• Probability of drawing 0 defective bulbs: To calculate this probability, we need to find the probability of drawing a non-defective bulb in all 4 draws. The probability of drawing a non-defective bulb in one draw is 1 - 1/5 = 4/5. Since the draws are independent, we can multiply the probabilities together: P(X = 0) = (4/5)^4 ≈ 0.4096

• Probability of drawing 1 defective bulb: To calculate this probability, we need to find the probability of drawing exactly 1 defective bulb in the 4 draws. This can happen in 4 different ways: DNNN, NDNN, NNDN, NNND, where D represents a defective bulb and N represents a non-defective bulb. The probability of each of these outcomes is (1/5) * (4/5)^3. Since there are 4 possible outcomes, we can multiply this probability by 4: P(X = 1) = 4 * (1/5) * (4/5)^3 ≈ 0.4096

• Probability of drawing 2 defective bulbs: Similarly, we can calculate the probability of drawing exactly 2 defective bulbs in the 4 draws. This can happen in 6 different ways: DDNN, DNDN, DNND, NDDN, NDND, NNDN. The probability of each of these outcomes is (1/5)^2 * (4/5)^2. Since there are 6 possible outcomes, we can multiply this probability by 6: P(X = 2) = 6 * (1/5)^2 * (4/5)^2 ≈ 0.1536

• Probability of drawing 3 defective bulbs: This can happen in 4 different ways: DDDN, DDND, DNDD, NDDD. The probability of each of these outcomes is (1/5)^3 * (4/5). Since there are 4 possible outcomes, we can multiply this probability by 4: P(X = 3) = 4 * (1/5)^3 * (4/5) ≈ 0.0512

• Probability of drawing 4 defective bulbs: This can only happen in 1 way: DDDD. The probability of this outcome is (1/5)^4: P(X = 4) = (1/5)^4 ≈ 0.0016

Therefore, the probability distribution of the number of defective bulbs in the sample is as follows: P(X = 0) ≈ 0.4096 P(X = 1) ≈ 0.4096 P(X = 2) ≈ 0.1536 P(X = 3) ≈ 0.0512 P(X = 4) ≈ 0.0016