Five men and four women are waiting to be interviewed for jobs. If they are all selected in random​ order, find the probability that no man will be interviewed until at least two women have been interviewed.

To find the probability that no man will be interviewed until at least two women have been interviewed, we need to consider the different possible arrangements of the candidates.

Let's break down the problem into separate cases:

Case 1: The first two candidates interviewed are women.
In this case, there are 4 women remaining and 5 men remaining to be interviewed.
The number of ways to arrange the remaining candidates is (4 women + 5 men)!, since all candidates are considered distinct.
So, for this case, the total number of possible arrangements is (4 + 5)! = 9!.

Case 2: The first candidate interviewed is a woman, and the second candidate interviewed is a man, followed by a woman.
In this case, there are 3 women remaining and 5 men remaining to be interviewed.
The number of ways to arrange the remaining candidates is (3 women + 5 men)!
So, for this case, the total number of possible arrangements is (3 + 5)! = 8!.

Case 3: The first candidate interviewed is a woman, and the second candidate interviewed is also a woman, followed by a man.
In this case, there are 3 women remaining and 4 men remaining to be interviewed.
The number of ways to arrange the remaining candidates is (3 women + 4 men)!
So, for this case, the total number of possible arrangements is (3 + 4)! = 7!.

Since there are three possible cases, we need to sum the total number of possible arrangements for each case:

Total number of possible arrangements = (9! + 8! + 7!)

To find the probability, we divide the favorable outcomes (total number of possible arrangements) by the total number of outcomes, which is the number of ways to arrange all 9 candidates:

Total number of outcomes = 9!

Finally, the probability is:

Probability = (9! + 8! + 7!) / 9!

To find the probability that no man will be interviewed until at least two women have been interviewed, we need to consider the different arrangements of the men and women.

First, let's determine the total number of possible arrangements of all nine candidates. Since there are nine candidates in total, we have 9! (9 factorial) ways to arrange them.

Now, let's consider the favorable arrangements where no man is interviewed until at least two women have been interviewed. In these arrangements, there must be at least two women interviewed before any man is interviewed.

To calculate this, we will break it down into cases:

Case 1: Exactly two women are interviewed before any man.
The arrangement for this case would be two women followed by any arrangement of the remaining seven candidates (which include three men and two remaining women). So, we have 2! * 7! arrangements for this case.

Case 2: Exactly three women are interviewed before any man.
The arrangement for this case would be three women followed by any arrangement of the remaining six candidates (which include two men and one remaining woman). So, we have 3! * 6! arrangements for this case.

Case 3: Four women are interviewed before any man.
The arrangement for this case would be four women followed by any arrangement of the remaining five candidates (which include one man). So, we have 4! * 5! arrangements for this case.

Now, let's calculate the total number of favorable arrangements by summing up the arrangements for each case:

Total number of favorable arrangements = (2! * 7!) + (3! * 6!) + (4! * 5!)

Finally, we can calculate the probability by dividing the total number of favorable arrangements by the total number of possible arrangements:

Probability = Total number of favorable arrangements / Total number of possible arrangements = [(2! * 7!) + (3! * 6!) + (4! * 5!)] / 9!

We must exclude the cases where the

first two are MW and MM
with no restriction, the number of ways to arrange the interviews is
9!/(4!5!) = 126
case of MM = 7!/(3!4!) = 35
case of MF = 7!/(4!3!) = 35
Prob(your stated event) = 1 - 70/126 = 4/9

answer ' =55