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31. From a lot of 30 bulbs which include 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.
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Q: 31. From a lot of 30 bulbs which include 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.

Question Analysis & Solution

Step-by-Step Solution

  1. Define the random variable:

    Let X be the number of defective bulbs in the sample of 2.

    X can take values 0, 1, or 2.

  2. Calculate the probabilities:

    Total number of bulbs = 30

    Number of defective bulbs = 6

    Number of non-defective bulbs = 30 - 6 = 24

    Since the bulbs are drawn with replacement, the probabilities remain constant for each draw.

    1. P(X = 0): Probability of getting 0 defective bulbs (both non-defective)

      P(X = 0) = (24/30) * (24/30) = (4/5) * (4/5) = 16/25

    2. P(X = 1): Probability of getting 1 defective bulb

      This can happen in two ways: Defective then Non-defective OR Non-defective then Defective

      P(X = 1) = (6/30) * (24/30) + (24/30) * (6/30) = 2 * (1/5) * (4/5) = 8/25

    3. P(X = 2): Probability of getting 2 defective bulbs

      P(X = 2) = (6/30) * (6/30) = (1/5) * (1/5) = 1/25

  3. Probability Distribution:

    The probability distribution of X is:

    X P(X)
    0 16/25
    1 8/25
    2 1/25
  4. Calculate the mean:

    Mean (μ) = Σ [X * P(X)]

    μ = (0 * 16/25) + (1 * 8/25) + (2 * 1/25) = 0 + 8/25 + 2/25 = 10/25 = 2/5 = 0.4

Correct Answer: 0.4

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