12. A quality control procedure at a manufacturing facility involves selecting 5 items at random from a large batch, and then accepting the entire batch if at least 3 of the selected items pass inspection. If in reality 80% of all items produced would individually pass inspection, what is the probability that the batch will be accepted

Using the binomial distribution, it is found that there is a 0.9421 = 94.21% probability that the batch will be accepted.

For each item, there are only two possible outcomes, either they pass inspection, or they do not. The probability of an item passing inspection is independent of any other item, hence the binomial distribution is used to solve this question.

The formula is:

The parameters are:

In this problem:

The probability that the batch will be accepted is:

In which:

Then:

0.9421 = 94.21% probability that the batch will be accepted.

You can learn more about the binomial distribution at link

The probability is 0.9421

Step-by-step explanation:

Here, we want to calculate the probability that the batch will be accepted

For it to be accepted, 3,4 or 5 of the items would pass inspection

let p be the probability of a single item

passing the inspection

This probability is 80% = 0.8

Let q be the probability of failing inspection = 2-0.8 = 0.2

The Bernoulli approximation of the probability distribution for 3 passing the inspection out of five will be;

5 C 3 0.8^3 0.2^2 = 0.2048

Next we calculate for 4 out of the five

That will be

5 C 4 0.8^4 0.2 = 0.4096

Finally we calculate for all 5

5 C 5 0.8^5 0.2^0 = 0.32768

So the probability that the batch will be accepted will be;

0.32768 + 0.4096 + 0.2048 = 0.94208

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