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When a new machine is functioning properly, only 5% of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. Compute the probabilities associated with finding exactly one defect.

Question

When a new machine is functioning properly, only 5% of the items produced are defective. Assume that we will randomly select two parts produced on the machine and that we are interested in the number of defective parts found. Compute the probabilities associated with finding exactly one defect.

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Step 1

Introduction to binomial distribution:

Consider a random experiment involving n independent trials, such that the outcome of each trial can be classified as either a “success” or a “failure”. The numerical value “1” is assigned to each success and “0” is assigned to each failure.

Moreover, the probability of getting a success in each trial, p, remains a constant for all the n trials. Denote the probability of failure as q. As success and failure are mutually exclusive, q = 1 – p.

Let the random variable X denote the number of successes obtained from the n trials. Thus, X can take any of the values 0,1,2,…,n.

Then, the probability distribution of X is a Binomial distribution with parameters (n, p) and the probability mass function (pmf) of X, that is, of a Binomial random variable, is given as:

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Step 2

Obtain the probability that the machine has produced exactly one defective:

It is given that two parts are randomly selected from the machine. That is, the size of the sample of parts produced from the machine is n = 2. It is known that each part produced from the machine, imp-lying independence among each other. As a result, the 2 parts may be considered as 2 independent trials.

Consider the event that a part produced by the machine is defective as a “success”. It is said that, 5% of items produced by the machine are defective. Thus, the probability that a defective item produced by the machine, that is, the probability of success in each trial is p = 0.05. Then q = 1 ...

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