Question

Four engineers, A, B, C, and D, have been scheduled for job interviews at 10 a.m. on Friday, January 13, at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms 1, 2, 3, and 4, respectively. However, the manager's secretary does not know this, so assigns them to the four rooms in a completely random fashion (what elsel). (a) What is the probability that all four end up in the correct rooms? (Round your answer to four decimal places.) (b) What is the probability that none of the four ends up in the correct room?

Answer 1

The probability that all four engineers end up in the correct rooms is about 0.0417, while the probability that none of the engineers end up in their correct rooms, known as a derangement, is approximately 0.375.

Step-by-step explanation:

The probability questions described involve a branch of mathematics known as combinatorics, and more specifically, they involve the concept of permutations and a special permutation called a derangement.

Probability of All Engineers in Correct Rooms

To find the probability that all four engineers end up in their pre-assigned rooms, we consider the fact that there is only one way for this to happen out of the total possible permutations of the four engineers. Since there are 4! (factorial of 4) ways to arrange four items, the probability is:

Probability = 1 / 4! = 1 / 24 or approximately 0.0417 (rounded to four decimal places).

Probability of No Engineers in Correct Rooms

The probability that none of the four engineers end up in their correct rooms is a classical example of a derangement problem. The number of derangements for four items is 9. Therefore, the probability in this case is:

Probability = Number of derangements / 4! = 9 / 24 or approximately 0.375 (rounded to four decimal places).