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12. A human resources director for a large corporation claims the probability a randomly
selected employee arrives to work on time is 0.95.
(a) If we take a random sample of 3 employees, what's the probability that all of them arrive on
time?
(b) If we take a random sample of 3 employees, what's the probability that none of them arrive
on time?
(c) If we take a random sample of 3 employees, what's the probability that at least one of them
arrives on time?
I

Respuesta :

Using the binomial distribution, it is found that the probabilities are given as follows:

a) 0.857375 = 85.7375%.

b) 0.000125 = 0.0125%.

c) 0.999875 = 99.9875%.

What is the binomial distribution formula?

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem, the values of the parameters are given as follows:

n = 3, p = 0.95.

Item a:

The probability is P(X = 3), hence:

P(X = 3) = (0.95)^3 = 0.857375 = 85.7375%.

Item b:

The probability is P(X = 0), hence:

P(X = 0) = (0.05)^3 = 0.000125 = 0.0125%.

Item c:

The probability is P(X > 0), hence:

P(X > 0) = 1 - P(X = 0) = 1 - 0.000125 = 0.999875 = 99.9875%.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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