During a manufacturing process 15 units are randomly selected each day from the production line to check the percent defective. from the historical information it is known that the probability of a defective unit is 0.05. any time that two or more defectives are found in the sample of 15, the process is stopped. this procedure is used to provide a signal in case the probability of a defective is increased. what is the proability that on any given day the production process will be stopped?

There is a 17.09% proability that on any given day the production process will be stopped

Step-by-step explanation:

The production will be stopped if there are two or more defective units, so we have to find .

This is . We know that either we have less than 2 defecive units, or we have at least 2 defective units. The sum of the probabilities is decimal 1. So

So, we have to find P(X < 2), that is:

Since either an unit is defective or it is not, that is, there are ony two possible outcomes, P(X = 0) and P(X = 1) are binomial probabilities.

Binomial probability

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

In which is the number of different combinatios of x objects from a set of n elements, given by the following formula.

And is the probability of X happening.

So:

The probability of a defective unit is 0.05, so

The size of the sample is 15, so

There is a 17.09% proability that on any given day the production process will be stopped

both equations and proportions are better.

to find the total number of combinations, multiply the quantity of each item by each other:

2 pairs of shoes x 3 pairs of shorts = 6 different combinations.