Statistics/Probability Exercises

kittykat16

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Hi, I have to complete this homework for Monday and while I've had relatively little trouble for the first few questions, I am entirely stuck on the last two :/ I haven't been able to work them out at all, any input or advice would be really appreciated!

1. A group of 2N boys and 2N girls is randomly divided into two equal groups. What is the probability that each group has the same number of boys and girls?

2. A committee of 5 people is randomly selected from 5 married couples (i.e. from 12 people). What is the probability that a husband and wife do not belong to that committee?

Thank you very much for your help!
 
#2

Hi, I have to complete this homework for Monday and while I've had relatively little trouble for the first few questions, I am entirely stuck on the last two :/ I haven't been able to work them out at all, any input or advice would be really appreciated!

2. A committee of 5 people is randomly selected from 5 married couples (i.e. from 12 people). What is the probability that a husband and wife do not belong to that committee?

Thank you very much for your help!
The first person selected can be any one of the 10:
......(10/10)

The 2nd person can be all but one of the remaining 9:
......(10/10)(8/9)

For the 3rd selection, there are 8 to choose from, but 2 are not allowed:
......(10/10)(8/9)(6/8)

Can you see the pattern?
 
The first person selected can be any one of the 10:
......(10/10)

The 2nd person can be all but one of the remaining 9:
......(10/10)(8/9)

For the 3rd selection, there are 8 to choose from, but 2 are not allowed:
......(10/10)(8/9)(6/8)

Can you see the pattern?

I see the pattern but I'm unsure of the next step to take :(

The 4th and 5th selection would be (10/10)(8/9)(6/8)(4/7)(2/6) but what does that mean for me?

I'm really sorry, I am utterly confused
 
#1

Hi, I have to complete this homework for Monday and while I've had relatively little trouble for the first few questions, I am entirely stuck on the last two :/ I haven't been able to work them out at all, any input or advice would be really appreciated!

1. A group of 2N boys and 2N girls is randomly divided into two equal groups. What is the probability that each group has the same number of boys and girls?

Suppose the first N selected are boys, and the following N are all girls. That is not very likely, but is ONE of many ways to wind up with N of each, boys and girls.

Use Combinations to find out how many ways to choose N of 2N boys.
......and how many ways to choose N of 2N girls.
How do you use those numbers to find the number of Combinations of BOTH boys and girls?

Find probability by comparing that number to the combinations of ways to select 2N out of the original 4N people.
 
I see the pattern but I'm unsure of the next step to take :(

The 4th and 5th selection would be (10/10)(8/9)(6/8)(4/7)(2/6) but what does that mean for me?

I'm really sorry, I am utterly confused
Those five fractions are the probabilities that the next randomly selected person will NOT be the spouse of somebody already selected. The product of the five is then the probability that NO husband/wife pair will be selected.

Where the question says "a husband and wife," another possible interpretation is that some specific pair is not on the committee. That would be much easier!
 
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