Question

Consider a shipment of 1000 items into a factory. Suppose the factory can tolerate about \(5 \%\) defective items. Let \(X\) be the number of defective items in a sample without replacement of size \(n=10 .\) Suppose the factory returns the shipment if \(X \geq 2\). (a) Obtain the probability that the factory returns a shipment of items that has \(5 \%\) defective items. (b) Suppose the shipment has \(10 \%\) defective items. Obtain the probability that the factory returns such a shipment. (c) Obtain approximations to the probabilities in parts (a) and (b) using appropriate binomial distributions. Note: If you do not have access to a computer package with a hypergeometric command, obtain the answer to (c) only. This is what would have been done in practice 20 years ago. If you have access to \(\mathrm{R}\), then the command dhyper \((\mathrm{x}, \mathrm{D}, \mathrm{N}-\mathrm{D}, \mathrm{n})\) returns the probability in expression (3.1.7).

Short answer

The solutions can be found by computing the above-provided formulas. For exact values, it is recommended to use a programmable software or calculator to perform the necessary numerical computations.

Step-by-step solution

Calculate Probability for Part (a)

The fraction of shipment that is defective is \(5\%\) i.e., \(D = 0.05 * 1000 = 50\). Hence, we can write the probability that defective items in the sample are greater or equal to 2 as: \[P(X \geq 2) = 1 - P(X=0) - P(X=1)\]\[P(X \geq 2) = 1 - \frac{\binom{50}{0}*\binom{950}{10}}{\binom{1000}{10}} - \frac{\binom{50}{1}*\binom{950}{9}}{\binom{1000}{10}}\]

Calculate Probability for Part (b)

With \(10\%\) defective items in the shipment, \(D = 0.1 * 1000 = 100\). Thus, we can state the probability that defective items in the sample are greater or equal to 2 as: \[P(X \geq 2) = 1 - P(X=0) - P(X=1)\]\[P(X \geq 2) = 1 - \frac{\binom{100}{0}*\binom{900}{10}}{\binom{1000}{10}} - \frac{\binom{100}{1}*\binom{900}{9}}{\binom{1000}{10}}\]

Approximate Probabilities using Binomial Distribution for Part (c)

The binomial approximation of the hypergeometric distribution can be applied when \(N\) is large and \(D\) is small compared to \(N\). Approximate the probability of returning a shipment with \(5\%\) and \(10\%\) defective items by substituting the binomial probabilities in the hypergeometric probability function: For Part (a): \[P(X \geq 2) = 1 - P(X=0) - P(X=1)\]\[P(X \geq 2) = 1 - \binom{10}{0}*(0.05)^{0}*(0.95)^{10} - \binom{10}{1}*(0.05)^{1}*(0.95)^{9}\]For Part (b): \[P(X \geq 2) = 1 - P(X=0) - P(X=1)\]\[P(X \geq 2) = 1 - \binom{10}{0}*(0.1)^{0}*(0.9)^{10} - \binom{10}{1}*(0.1)^{1}*(0.9)^{9}\]