if 3 couples sit in a row of chairs what is the probability that each couple sits together?
what is the probability all the women sit together and the men may or may not be in the group?
what is the probability that the women sit together and the men sit together?
That depends upon the couples.
How is this a statistics question??
probability , i guess... its for a stats class.
im pretty sure my teacher doesnt word the questions very well.
In addition to knowing nothing about the couples, we don't know the event. Couples usually sit together in church, but often sit with their own gender in other situations. At a party, the couples often mingle and don't stay together.
You could tell the teacher that the question can't be answered.
a) consider each couple as one unit, so you are just arranging 3 units.
Number of ways = 3! = 6
b) I don't understand the question
c) for the men to sit together, we could have
MMMWWW ****
WMMMWW
WWMMMW
WWWMMM
Number of ways to do ***
= 3x2x1x3x2x1=36
but each of the other calculations are the same
number of ways = 4x6 = 24
Last line of previous post should have been:
number of ways = 4x36 = 144
To find the probabilities in these scenarios, we need to consider the total number of possible seating arrangements and then count the number of favorable arrangements that meet the given conditions. Let's analyze each scenario separately:
1. Probability that each couple sits together:
We have 3 couples, which means there are 6 people to be seated. We can treat each couple as a single unit, reducing the problem to permuting 3 units (couples). The total number of possible arrangements is 3! = 3 * 2 * 1 = 6. However, for each couple, they can arrange themselves within their unit in 2 ways (husband on the left or right of the wife). So, the total number of favorable arrangements where each couple sits together is 2^3 = 8. The probability is then 8/6 = 4/3. However, probabilities must be between 0 and 1, so we cap it at 1. Therefore, the probability that each couple sits together is 1.
2. Probability that all the women sit together, and the men may or may not be in the group:
In this case, since all the women must sit together, we can treat them as a single unit. So, we have 4 units: the group of women and the three individual men. The total number of possible arrangements is 4!. However, within the group of women, they can arrange themselves in 3! ways. This accounts for their relative positions. Therefore, the total number of favorable arrangements is 3! * 3!. The probability is then (3! * 3!) / 4! = 1/4 = 0.25.
3. Probability that the women sit together, and the men sit together:
For this case, both the women and the men must sit together. We can treat each gender as a single unit. So, we have 2 units: the group of women and the group of men. The total number of arrangements is 2!. However, within each group, the individuals can arrange themselves. Thus, we have (3! * 3!) favorable arrangements. The probability is (3! * 3!) / 2! = 6/2 = 3.
To summarize:
1. Probability that each couple sits together: 1
2. Probability that all the women sit together, and the men may or may not be in the group: 0.25
3. Probability that the women sit together, and the men sit together: 3