Answer to Question #105546 in Statistics and Probability for George Mwila Chola

Question

In a food production process, packaged items are sampled as they come off a production line: a random sample of 5 items from each production batch is checked to see if each is tightly parked. A batch is accepted if all 5 sample items are satisfactory, and rejected if there are 3 or more unsatisfactory packages in it, otherwise a further sample is taken before making a decision. If in fact the packing machine is giving 80% of items properly parked, what is the probability that this second sample will be necessary? The second sample also consists of 5 items. What is the probability that, out of the two samples (10 items), there are 9 satisfactory? What are the assumptions behind your calculations?

Answer 1

The second sample will be necessary:

The packing machine is giving 80%=0.8 of items properly parked, unsatisfactory packages 1-0.8=0.2.

3 or more unsatisfactory packages: 3 or 4 or 5

Choose 3 from 5 unsatisfactory packages:

$p_1=\frac{3}{5}\cdot 0.2$

Choose 4 from 5 unsatisfactory packages:

$p_2=\frac{4}{5}\cdot0.2$

Choose 5 from 5 unsatisfactory packages:

$p_3=\frac{5}{5}\cdot0.2$

$P=p_1+p_2+p_3=\frac{3}{5}\cdot0.2+\frac{4}{5}\cdot0.2+\frac{5}{5}\cdot0.2=0.48$

The probability 0.48 that second sample will be necessary.

Out of the two samples (10 items), there are 9 satisfactory:

We use the Bernoulli formula:

$P=(\begin{matrix} k \\ n \end{matrix})p^{k}q^{n-k}$

In our case

$p=0.8\\ q=0.2\\ ( \begin{matrix} 9\\ 10 \end{matrix})=\frac{10!}{9!(10-9)!}=\frac{10\cdot 9!}{9!\cdot 1!}=10\\ P=10\cdot 0.8^9\cdot0.2^{10-9}=10\cdot 0.8^9\cdot0.2^1\approx 0.268$

The 0.268 probability that, out of the two samples (10 items), there are 9 satisfactory