A corporation has 10 manufacturing plants. Of​ these, seven are domestic and three are outside of the country. Each year a performance evaluation is conducted for four randomly selected plants. What is the probability that a performance evaluation will include at least one plant outside of the​ country? (Hint: Begin by finding the probability that only domestic plants are​ selected.)

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Answer:

The probability that a performance evaluation will include at least one plant outside of the​ country is 0.76.

Step-by-step explanation:

The random variable X can be defined as the number of outside of the country plants that are selected for a performance evaluation.

A sample of n = 4 plants are selected each year for a performance evaluation.

The probability of selecting an outside of the country plant is, p = 0.30.

A plant selected can either be a domestic plant or outside of the country plant. They are not dependent on each other.

Thus, the random variable X follows a Binomial distribution with parameters n = 4 and p = 0.30.

The probability mass function of X is:

[tex]P(X=x)={4\choose x}\ 0.30^{x}(1-0.30)^{4-x};\ x=0,1,2,3...[/tex]

Compute the probability that a performance evaluation will include at least one plant outside of the​ country as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             [tex]=1-{4\choose 0}\ 0.30^{0}(1-0.30)^{4-0}\\\\=1-(1\times 1\times 0.2401)\\\\=1-0.2401\\\\=0.7599\\\\\approx 0.76[/tex]

Thus, the probability that a performance evaluation will include at least one plant outside of the​ country is 0.76.

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