You have been hired as the quality control office of a pharmaceutical company that manufactures aspirin tablets. According to the quality assurance protocol, your job is to do the following. For each shipment of possibly thousands and thousands of aspirin tablets, randomly select and test 24 tablets, then accept the batch if there is only one or no tablet that doesn’t meet the test specification. If, at the pharmaceutical company, the probability that randomly selected tablet is defective is .04, what is the probability that a randomly selected shipment of 24 tablets will be accepted?

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

[tex]X \sim Binom(n=24, p=0.04)[/tex]

On this case p=0.04 represent the probability that a random the tablet would be NOT accepted

The probability mass function for the Binomial distribution is given as:

[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]

Where (nCx) means combinatory and it's given by this formula:

[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]

They accept the batch if there is only one or no tablet that doesn’t meet the test specification, so let's find the individual probabilities for this case:

[tex]P(X=0)=(24C0)(0.04)^0 (1-0.04)^{24-0}=0.375413[/tex]

[tex]P(X=1)=(24C1)(0.04)^1 (1-0.04)^{24-1}=0.[/tex]

So then the probability that the batch would be accepeted is:

[tex] 0.375413+0.375413=0.7508[/tex]