Question

A human resources director for a large corporation claims the probability a randomly selected employe 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?

Answer 1

The probability that all three employees arrive on time is 0.857375 or 85.74%.

The probability that none of the employees arrive on time is 0.142625 or 14.26%.

The probability that at least one employee arrives on time is 0.857375 or 85.74%.

Step-by-step explanation:

(a) To find the probability that all three employees arrive on time, we can multiply the individual probabilities together.

Since the probability of each employee arriving on time is 0.95, the probability that all three employees arrive on time is 0.95 x 0.95 x 0.95

= 0.857375 or 85.74%.

(b) To find the probability that none of the employees arrive on time, we can simply subtract the probability of all employees arriving on time from 1.

So, the probability that none of the employees arrive on time is 1 - 0.857375

= 0.142625 or 14.26%.

(c) To find the probability that at least one employee arrives on time, we can subtract the probability of none of the employees arriving on time from 1.

So, the probability that at least one employee arrives on time is 1 - 0.142625

= 0.857375 or 85.74%.