810) See Ex. 809. a) Find the probability of obtaining 2 nonconforming units on the first sample of 2 and 1 nonconforming unit on the second sample of 2. b) What is the probability of 0 nonconforming units on the first sample and 1 nonconforming units on the second? The first sample is returned before the second sample is selected.   ex 809 It is given that, the probability of obtaining 1 nonconforming unit in a sample of two is 0.39 and the probability of 2 nonconforming units in a sample of two is 0.16. Consider a random variable x that defines the number of nonconforming unit in the sample. Hence, according to question, P(X = 1) = 0.39 and P(X = 2) = 0.16. It is known that the sum of the probabilities is 1. That is, P(X = 0) + P(X = 1) + P(X = 2) = 1 P(X = 0) = 1 – P(X = 1) – P(X = 2) = 1 – 0.39 – 0.16 = 0.45. Thus, the probability of zero nonconforming units in a sample of two is 0.45.

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810) See Ex. 809. a) Find the probability of obtaining 2 nonconforming units on the first sample of 2 and 1 nonconforming unit on the second sample of 2. b) What is the probability of 0 nonconforming units on the first sample and 1 nonconforming units on the second? The first sample is returned before the second sample is selected.

 

ex 809

It is given that, the probability of obtaining 1 nonconforming unit in a sample of two is 0.39 and the probability of 2 nonconforming units in a sample of two is 0.16.

Consider a random variable x that defines the number of nonconforming unit in the sample.

Hence, according to question, P(X = 1) = 0.39 and P(X = 2) = 0.16.

It is known that the sum of the probabilities is 1.

That is, P(X = 0) + P(X = 1) + P(X = 2) = 1

P(X = 0) = 1 – P(X = 1) – P(X = 2) = 1 – 0.39 – 0.16 = 0.45.

Thus, the probability of zero nonconforming units in a sample of two is 0.45.

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