it is known that 5% of all parts manufactured at Plant F are defective. If 30 part are randomly selected what is the probability that all 30 parts are not defective and what is the probability that at least one of the 30 parts is defective?

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To solve this problem, we need to use the concept of probability and the binomial distribution.

1. Probability that all 30 parts are not defective:
The probability of selecting a non-defective part from Plant F is 1 - 0.05 = 0.95 (since the defective rate is 5% or 0.05). We can use the binomial probability formula to calculate the probability of selecting all 30 non-defective parts:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where:
- P(X=k) is the probability of selecting exactly k non-defective parts
- C(n, k) is the number of ways to choose k parts out of n parts (combination)
- p is the probability of selecting a non-defective part (0.95)
- n is the number of parts selected (30)
- k is the number of non-defective parts (30)

So, for this case:
P(all 30 parts are not defective) = P(X = 30) = C(30, 30) * (0.95)^30 * (1-0.95)^(30-30)

The combination formula C(n, k) = n! / (k! * (n-k)!), where n! represents n factorial.

2. Probability that at least one of the 30 parts is defective:
To find the probability that at least one part is defective, we can calculate the complement of the probability that none of the parts are defective. In other words:

P(at least one defective) = 1 - P(no defective)

P(no defective) = P(all 30 parts are not defective)

Therefore,
P(at least one defective) = 1 - P(X = 30)

Now, you can plug these values into a calculator or use statistical software to evaluate these probabilities.

Note: Alternatively, you can use a calculator with the binomial distribution function to directly calculate P(X = k) and P(X ≥ k), where k represents the number of non-defective parts.