Question

A pharmaceutical company receives large shipments of ibuprofen tablets and uses an acceptance sampling plan. This plan randomly selects and tests 26 tablets, then accepts the whole batch if there is at most one that doesn't meet the required specifications. What is the probability that this whole shipment will be accepted if a particular shipment of thousands of ibuprofen tablets actually has a 4% rate of defects

Answer 1

Answer:

0.7208 = 72.08% probability that this whole shipment will be accepted.

Step-by-step explanation:

For each tablet, there are only two possible outcomes. Either it meets the required specifications, or it does not. The probability of a tablet meeting the required specifications is independent of any other tablet, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

4% rate of defects

This means that $p = 0.04$

26 tablets

This means that $n = 26$

What is the probability that this whole shipment will be accepted?

Probability that at most one tablet does not meet the specifications, which is:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Thus

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{26,0}.(0.04)^{0}.(0.96)^{26} = 0.3460$

$P(X = 1) = C_{26,1}.(0.04)^{1}.(0.96)^{25} = 0.3748$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.3460 + 0.3748 = 0.7208$

0.7208 = 72.08% probability that this whole shipment will be accepted.