3.4k views
4 votes
A human resources director for a large corporation claims the probability a randomly selected employee arrives to work on time is .92

1 Answer

5 votes

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%.

Here's the breakdown of the probabilities for each scenario:

(a) All 3 employees on time:

Since each arrival is an independent event, we simply multiply the individual probabilities: 0.95 (on-time) x 0.95 x 0.95 = 0.8574. So, there's an 85.74% chance all three will be punctual.

(b) None of the 3 employees on time:

Again, using independence and the probability of being late (1 - 0.95 = 0.05), we get 0.05 x 0.05 x 0.05 = 0.000125. This translates to a meager 0.0125% chance of everyone being late.

(c) At least one employee on time:

This is the easiest one! We simply take the opposite of the "none on time" scenario: 1 - 0.000125 = 0.999875. Therefore, there's a whopping 99.99% chance that at least one employee will be on time.

Question:

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?

User Vineet Kasat
by
8.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories