Question

A company is interested in evaluating its current
inspection procedure for shipments of 50 identical
items. The procedure is to take a sample of 5 and
pass the shipment if no more than 2 are found to be
defective. What proportion of shipments with 20% defectives
will be accepted?

Answer 1

Answer on Question #72628, Math / Statistics and Probability

A company is interested in evaluating its current inspection procedure for shipments of 50 identical items. The procedure is to take a sample of 5 and pass the shipment if no more than 2 are found to be defective. What proportion of shipments with 20% defectives will be accepted?

Solution

Let $X$ be the number of defective items in the sample of $n = 5$ taken from $N = 50$ without replacement. Suppose the shipment has $M$ defective items. Thus $X$ is a hypergeometric variable with $X \sim \text{Hypergeometric}(n = 5, N = 50, M = pN)$, where $p$ is the proportion of defectives in the shipment. If $p = 0.2$ then $M = 10$.

$$P (X = k) = h _ {(M, n, N)} (k) = \frac {\binom {M} {k} \binom {N - M} {n - k}}{\binom {N} {n}}$$

The probability of accepting the shipment is

$$\begin{array}{l} P (X \leq 2) = P (X = 0) + P (X = 1) + P (X = 2) = \\ = h _ {(1 0, 5, 5 0)} (0) + h _ {(1 0, 5, 5 0)} (1) + h _ {(1 0, 5, 5 0)} (2) = \\ = \frac {\binom {1 0} {0} \binom {5 0 - 1 0} {5 - 0}}{\binom {5 0} {5}} + \frac {\binom {1 0} {1} \binom {5 0 - 1 0} {5 - 1}}{\binom {5 0} {5}} + \frac {\binom {1 0} {2} \binom {5 0 - 1 0} {5 - 2}}{\binom {5 0} {5}} \approx \\ \approx 0. 3 1 0 5 6 + 0. 4 3 1 3 4 + 0. 2 0 9 8 4 \approx 0. 9 5 1 7 \\ \end{array}$$ $$\begin{array}{l} \frac {\binom {1 0} {0} \binom {5 0 - 1 0} {5 - 0}}{\binom {5 0} {5}} = \frac {\frac {1 0 !}{0 ! (1 0 - 0) !} \cdot \frac {4 0 !}{5 ! (4 0 - 5) !}}{\frac {5 0 !}{5 ! (5 0 - 5) !}} = \frac {1 \cdot \frac {4 0 (3 9) (3 8) (3 7) (3 6)}{1 (2) (3) (4) (5)}}{\frac {5 0 (4 9) (4 8) (4 7) (4 6)}{1 (2) (3) (4) (5)}} = \\ = 0. 3 1 0 5 6 \\ \end{array}$$ $$\begin{array}{l} \frac {\binom {1 0} {1} \binom {5 0 - 1 0} {5 - 1}}{\binom {5 0} {5}} = \frac {\frac {1 0 !}{1 ! (1 0 - 1) !} \cdot \frac {4 0 !}{4 ! (4 0 - 4) !}}{\frac {5 0 !}{5 ! (5 0 - 5) !}} = \frac {1 0 \cdot \frac {4 0 (3 9) (3 8) (3 7)}{1 (2) (3) (4)}}{\frac {5 0 (4 9) (4 8) (4 7) (4 6)}{1 (2) (3) (4) (5)}} = \\ = 0. 4 3 1 3 4 \\ \end{array}$$ $$\begin{array}{l} \frac {\binom {1 0} {2} \binom {5 0 - 1 0} {5 - 2}}{\binom {5 0} {5}} = \frac {\frac {1 0 !}{2 ! (1 0 - 2) !} \cdot \frac {4 0 !}{3 ! (4 0 - 3) !}}{\frac {5 0 !}{5 ! (5 0 - 5) !}} = \frac {\frac {1 0 (9)}{2} \cdot \frac {4 0 (3 9) (3 8)}{1 (2) (3)}}{\frac {5 0 (4 9) (4 8) (4 7) (4 6)}{1 (2) (3) (4) (5)}} = \\ = 0. 2 0 9 8 4 \\ \end{array}$$

Answer: $\approx 0.9517; 95.17\%$.

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