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Distribution of defectives drawn from ten

Short answer

Drawing three from 10 items with four defective gives hypergeometric probabilities P(0)=1/6, P(1)=1/2, P(2)=3/10, P(3)=1/30. Mean is 1.2 and variance is 0.56.

Four defective items are mixed with six good items. Three items are drawn at random. Obtain the probability distribution for the number of defective items in the drawn sample. Hence obtain mean and variance for the distribution.

There are 10 items: 4 defective and 6 good. Three are drawn without replacement. Let X = number of defectives in the sample; X is hypergeometric (N = 10, K = 4, n = 3):

P(X = x) = C(4, x) · C(6, 3 − x) / C(10, 3), C(10, 3) = 120.

x   C(4,x)·C(6,3-x)     P(x)
0   1 × 20 = 20         20/120 = 1/6  = 0.1667
1   4 × 15 = 60         60/120 = 1/2  = 0.5000
2   6 × 6  = 36         36/120 = 3/10 = 0.3000
3   4 × 1  = 4          4/120  = 1/30 = 0.0333
Sum                     1

Mean

E(X) = Σ x P(x) = 0 + 0.5 + 0.6 + 0.1 = 1.2 (Check: nK/N = 3 × 4/10 = 1.2.)

Variance

E(X²) = 0 + 1(0.5) + 4(0.3) + 9(0.0333) = 0.5 + 1.2 + 0.3 = 2.0

Var(X) = E(X²) − [E(X)]² = 2.0 − 1.44 = 0.56

(Check: n(K/N)(1 − K/N)(N − n)/(N − 1) = 3 × 0.4 × 0.6 × 7/9 = 0.56.)

Answer: distribution as above, mean = 1.2, variance = 0.56

✓ Answer checked by Claude Sonnet 5.5

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