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?