A machine produces 0, 1 or 2 defective pieces in a day with associated probability of 1/6, 2/3 and 1/6, respectively. The mean value and the variance of the number of defective piecesproduced by the machine in a day, respectively, area)1 and 1/3b)1/3 and 1c)1 and 4/3d)1/3 and 4/3Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Mean value of the number of defective pieces:
To find the mean value of the number of defective pieces produced by the machine in a day, we need to multiply the number of defective pieces by their respective probabilities and sum them up.

Let X be the number of defective pieces. We have:
P(X = 0) = 1/6 (probability of producing 0 defective pieces)
P(X = 1) = 2/3 (probability of producing 1 defective piece)
P(X = 2) = 1/6 (probability of producing 2 defective pieces)

The mean value (μ) is given by the formula:
μ = Σ(X * P(X))

Calculating this:
μ = (0 * 1/6) + (1 * 2/3) + (2 * 1/6)
μ = 0 + 2/3 + 2/6
μ = 0 + 4/6 + 2/6
μ = 6/6
μ = 1

Therefore, the mean value of the number of defective pieces produced by the machine in a day is 1.

Variance of the number of defective pieces:
To find the variance of the number of defective pieces produced by the machine in a day, we need to calculate the squared deviations from the mean.

The variance (σ^2) is given by the formula:
σ^2 = Σ((X - μ)^2 * P(X))

Calculating this:
σ^2 = ((0 - 1)^2 * 1/6) + ((1 - 1)^2 * 2/3) + ((2 - 1)^2 * 1/6)
σ^2 = (1^2 * 1/6) + (0^2 * 2/3) + (1^2 * 1/6)
σ^2 = 1/6 + 0 + 1/6
σ^2 = 2/6
σ^2 = 1/3

Therefore, the variance of the number of defective pieces produced by the machine in a day is 1/3.

Conclusion:
The mean value and the variance of the number of defective pieces produced by the machine in a day are 1 and 1/3, respectively. Thus, the correct answer is option A: 1 and 1/3.

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