Three men and three women are waiting to be interviewed for jobs. If they are all selected in random order find the probability of each event. No man will be interviewed until at least two women have been interviewed.
1/20
Well, let's see. According to your question, we have three men and three women waiting to be interviewed for jobs. The order in which they are selected is random. We want to find the probability of each event.
Event A: No man will be interviewed until at least two women have been interviewed.
Event B: The women are interviewed before the men.
To calculate these probabilities, we need to consider the total number of possible orders and the number of favorable outcomes for each event.
First, let's determine the total number of possible orders. Since there are six people in total, the number of possible orders is 6! (factorial), which equals 720.
For Event A, we need to consider the different arrangements where no man is interviewed until at least two women have been interviewed. This means that the first two positions can only be women, and the remaining four positions can be any gender.
The number of favorable outcomes for Event A is then 3! * 4! (3 factorial multiplied by 4 factorial). This is because there are three women who can be chosen to fill the first two positions and then four remaining positions that can be filled by anyone.
So, the probability of Event A occurring is 3! * 4! / 6! = 144 / 720 = 1/5.
Now let's move on to Event B, where the women are interviewed before the men.
For this event, the first three positions need to be filled by women, and the remaining three positions can be filled by anyone.
The number of favorable outcomes for Event B is 3! * 3! (3 factorial multiplied by 3 factorial). This is because there are three women who can be chosen to fill the first three positions, and then three remaining positions that can be filled by either men or women.
So, the probability of Event B occurring is 3! * 3! / 6! = 36 / 720 = 1/20.
So, the probability of Event A is 1/5, and the probability of Event B is 1/20.
To find the probability of each event, let's consider the different possible outcomes.
First, let's count the number of possible arrangements:
The total number of ways to arrange the three men and three women is 6! (since there are 6 people in total).
Now, let's consider the event where no man is interviewed until at least two women have been interviewed. In this case, we can have the following arrangements:
1. WWMMMW
2. WWMWMW
3. WWMWWM
4. WWWMWM
5. WWWMMW
6. WWWMWM
7. WWWWMN
So, there are 7 possible arrangements where no man is interviewed until at least two women have been interviewed.
The probability of this event happening is given by:
P(event) = number of favorable outcomes / total number of possible outcomes
P(no man is interviewed until at least two women have been interviewed) = 7 / 6!
Simplifying:
P(event) = 7 / (6 × 5 × 4 × 3 × 2 × 1) = 7 / 720 = 1 / 103 ≈ 0.00971 or 0.971%
Therefore, the probability of no man being interviewed until at least two women have been interviewed is approximately 0.00971 or 0.971%.
To find the probability of each event, we first need to determine the total number of possible outcomes (sample space) and the number of favorable outcomes for each event.
Let's consider the total number of possible outcomes (sample space). Since there are 6 people (3 men and 3 women) waiting to be interviewed, we can arrange them in 6! (6 factorial) ways. This means there are 6 x 5 x 4 x 3 x 2 x 1 = 720 possible outcomes.
Now, let's calculate the number of favorable outcomes for each event:
Event A: At least two women are interviewed before any man.
To calculate the number of favorable outcomes for this event, we can consider the positions of the women and men during the interviews. Since there must be at least two women interviewed before any man, we have two possible arrangements:
1. Women, Women, Man, Man, Man, Man
2. Women, Man, Women, Man, Man, Man
For the first arrangement, we have 3! ways to arrange the women and 3! ways to arrange the men. So, there are 3! x 3! = 6 x 6 = 36 favorable outcomes for this arrangement.
For the second arrangement, we have 3! ways to arrange the women and 3! ways to arrange the men. So, there are also 3! x 3! = 6 x 6 = 36 favorable outcomes for this arrangement.
Therefore, the total number of favorable outcomes for Event A is 36 + 36 = 72.
Event B: The women are interviewed before the men.
To calculate the number of favorable outcomes for this event, we need to consider the arrangement of women and men during the interviews. Since all women must be interviewed before the men, there is only one possible arrangement:
1. Women, Women, Women, Man, Man, Man
In this case, we have 3! ways to arrange the women and 3! ways to arrange the men. So, there are 3! x 3! = 6 x 6 = 36 favorable outcomes for this arrangement.
Therefore, the total number of favorable outcomes for Event B is 36.
Now, we can calculate the probability of each event by dividing the number of favorable outcomes by the total number of possible outcomes:
Event A: Probability = Favorable Outcomes / Total Outcomes = 72 / 720 = 1 / 10 = 0.1 (or 10%)
Event B: Probability = Favorable Outcomes / Total Outcomes = 36 / 720 = 1 / 20 = 0.05 (or 5%)
P(w) = 3/6
p(ww) = 3/6 * 2/5 = 1/5
so, the chance that 2 women in a row are chosen first is 1/5. That is the chance that no man will be interviewed 1st or 2nd.