Combinations/probability question

[Micky1964]

Hi all. I'm doing a counting/probability problem here.
I can see how the correct answer for part b is arrived at by counting the number of combinations for "2 girls" then counting the number of combinations for "3 girls" , and adding these results.
But I've attached my first way of approaching it, which is wrong. Wondering if anyone can tell me why my approach doesn't give the correct answer. I know it's probably something obvious that I'm missing.
Thanks

 

[Dr.Peterson]

Hi all. I'm doing a counting/probability problem here.
I can see how the correct answer for part b is arrived at by counting the number of combinations for "2 girls" then counting the number of combinations for "3 girls" , and adding these results.
But I've attached my first way of approaching it, which is wrong. Wondering if anyone can tell me why my approach doesn't give the correct answer. I know it's probably something obvious that I'm missing.
Thanks

This is a common type of overcounting error.

If you choose 2 girls, and then fill the third spot with anyone, you will obtain the same set in multiple ways. For example, you might choose Ann and Beth first, and the third is Cathy; or you might choose Ann and Cathy first, and Beth as the third; or Beth and Cathy, then Ann.

That's why you get 48 instead of 40. Specifically, you counted each of the 4 ways to choose 3 girls three times, making it 12 rather than 4, exactly accounting for the excess of 8.

 

[Micky1964]

View attachment 28915

View attachment 28914

This is a common type of overcounting error.

If you choose 2 girls, and then fill the third spot with anyone, you will obtain the same set in multiple ways. For example, you might choose Ann and Beth first, and the third is Cathy; or you might choose Ann and Cathy first, and Beth as the third; or Beth and Cathy, then Ann.

That's why you get 48 instead of 40. Specifically, you counted each of the 4 ways to choose 3 girls three times, making it 12 rather than 4, exactly accounting for the excess of 8.

Thanks so much for your prompt reply.

So if I'm understanding correctly, my method was counting all the possible permutations of 4 girls into 3 spots, rather than the possible selections (sets, where order doesn't matter) of 4 girls into 3 spots...and to get from permutations to selections, you divide the total number of possible permutations by the number of possible ways those 3 girls could be arranged (ie 3!).

So it's safest to approach these types of questions by taking the 2 cases separately and adding them? ie How many ways to choose 2 girls + how many ways to choose 3 girls?

Also, where you say :

"Specifically, you counted each of the 4 ways to choose 3 girls three times, making it 12 rather than 4, exactly accounting for the excess of 8.",

isn't there 4x3x2 ways to choose 4 girls into 3 spots?

Once again, thanks for your help. I'm new on here so hopefully I'll get better at expressing exactly what I mean and also will learn how to input the appropriate mathematical symbols.

Mick

 

[Dr.Peterson]

So if I'm understanding correctly, my method was counting all the possible permutations of 4 girls into 3 spots, rather than the possible selections (sets, where order doesn't matter) of 4 girls into 3 spots...and to get from permutations to selections, you divide the total number of possible permutations by the number of possible ways those 3 girls could be arranged (ie 3!).

Not exactly. (See below.) What your method counts (as part of the total) is the ways to assign 4 girls into a pair (chosen first) and one other (chosen without regard to gender), and leave one not chosen.

So it's safest to approach these types of questions by taking the 2 cases separately and adding them? ie How many ways to choose 2 girls + how many ways to choose 3 girls?

Probably. In combinatorics, I never say that one way is best, because there may be another that would be at least as efficient. In this case, you could also subtract ways to choose 0 or 1 girls from the total ways to choose 3 students.

isn't there 4x3x2 ways to choose 4 girls into 3 spots?

That's why it wasn't exactly permutations that you counted! The two chosen are not distinct spots.

 

[Micky1964]

Not exactly. (See below.) What your method counts (as part of the total) is the ways to assign 4 girls into a pair (chosen first) and one other (chosen without regard to gender), and leave one not chosen.


Probably. In combinatorics, I never say that one way is best, because there may be another that would be at least as efficient. In this case, you could also subtract ways to choose 0 or 1 girls from the total ways to choose 3 students.


That's why it wasn't exactly permutations that you counted! The two chosen are not distinct spots.

Great! Thanks again.

 

[sky2rain]

For this kind of problem I would always assume what happens if there are equal large qty of boys and girls. With 3 spots chosen you would be looking at a permutation with repetition which has 2^3=8 possible line ups basing on 2 attributes at each spot. This would check out since for each of 2 attributes’ probability to be picked is equal, each line up has (1/2)*(1/2)*(1/2) = 1/8 probability. Then you would try figure out just how many line ups satisfy the conditions of there are at least 2 girls. So the question becomes to choose 2 spots from spot1, spot2, spot3 to identify as girls: C(3,2)=3. Then you add that to all spots identified as girls which is C(3,3)=1. Plug in the probability you get (1/8)*3+(1/8)*1=4/8. Now back in your question since the probability for girls and boys are different as b prob=3/5, g prob=2/5, yet the spot identification doesn’t change. So 2 spot identified as girl and 1 spot identified as boy, the probability would be written as (2/5)*(2/5)*(3/5), and since C(3,2) doesn’t change, instead of using (1/8)*3 you use (2/5)*(2/5)*(3/5)*3. And then plus the case for all three girls which is (2/5)*(2/5)*(2/5)*1.

 

[sky2rain]

Another confusing point is that you would think a line up for boy girl boy and boy boy girl are essentially the same so permutation shouldn’t apply. Take note that combination reduces the essentially same cases in a permutation only when the number of attribute is bigger than the spot to take them. In this case you need to choose 3 spots out of 2 attributes that’s why you can’t reduce the permutation in a straightforward way.

 

[sky2rain]

And a permutation of 3 spots with 2 attributes require repetition. You can’t have a permutation of number of spots bigger than attributes without repetition.

 

[sky2rain]

If this question changes to choosing 4 balls out of 5 colors of balls then there would have another approach as if the spots don’t matter much, but it wouldn’t be simpler than this way.

 

[sky2rain]

Another confusing point is that you would think a line up for boy girl boy and boy boy girl are essentially the same so permutation shouldn’t apply. Take note that combination reduces the essentially same cases in a permutation only when the number of attribute is bigger than the spot to take them. In this case you need to choose 3 spots out of 2 attributes that’s why you can’t reduce the permutation in a straightforward way.

To put it in a simpler way, combination formula reduces essentially same cases in a permutation if there is no repetition. Since there are 3 spots with 2 attributes, there must be repetition in that permutation.

 

[Micky1964]

Thanks sky2rain! Some good info there.

 

[Micky1964]

For this kind of problem I would always assume what happens if there are equal large qty of boys and girls. With 3 spots chosen you would be looking at a permutation with repetition which has 2^3=8 possible line ups basing on 2 attributes at each spot. This would check out since for each of 2 attributes’ probability to be picked is equal, each line up has (1/2)*(1/2)*(1/2) = 1/8 probability. Then you would try figure out just how many line ups satisfy the conditions of there are at least 2 girls. So the question becomes to choose 2 spots from spot1, spot2, spot3 to identify as girls: C(3,2)=3. Then you add that to all spots identified as girls which is C(3,3)=1. Plug in the probability you get (1/8)*3+(1/8)*1=4/8. Now back in your question since the probability for girls and boys are different as b prob=3/5, g prob=2/5, yet the spot identification doesn’t change. So 2 spot identified as girl and 1 spot identified as boy, the probability would be written as (2/5)*(2/5)*(3/5), and since C(3,2) doesn’t change, instead of using (1/8)*3 you use (2/5)*(2/5)*(3/5)*3. And then plus the case for all three girls which is (2/5)*(2/5)*(2/5)*1.

I like this approach. I need to give it some more thought to fully understand it but thanks!

 

[Dr.Peterson]

I like this approach. I need to give it some more thought to fully understand it but thanks!

I hope you see that this person's answer is wrong. Don't spend too much time on it.

 

[Micky1964]

I hope you see that this person's answer is wrong. Don't spend too much time on it.

I haven't had time to sit down with it yet. Hopefully when I do, I'll be able to see the error in this method.
I never like to use a method without understanding why it works anyway.

 

[sky2rain]

I haven't had time to sit down with it yet. Hopefully when I do, I'll be able to see the error in this method.
I never like to use a method without understanding why it works anyway.

Yeah I was wrong since these are people, not balls of the same color. Let me find out where I did wrong.

 

[sky2rain]

Yeah my approach only works for problems like coin toss where there are only two attributes. Here you actually have 10 attributes and each is an individual that can only take one spot so there is no repetition involved. The analysis is straightforward: when you have 4 girls taking 2 spots, there are c4,2 ways of arranging them. And the last spot was taken by one of 6 boys so c4,2*c6,1. And when you have three girls taking all spot then c4,3. So total events for this RV is c4,2*c6,1+c4,3, divide by the sample space c10,3, that is the probability… what Micky did was that he counted the girls not taken in the spot among the 8 people chosen for the last spot, which is not reasonable since when you do c4,2 it accounts for all 4 girls in the two spots to be taken. Hence if you write c8,1, it must accounts for 8 specific people. If girls are letter noted as A-D, boys are letter noted as E-J, without a fixed rule (gender) to assign them, your 8 people is assigned by a dynamic rule, which obviously won’t work. Even if you assign a fixed rule, say C-J as the 8 people, then while choosing the 2 girls out of 4, It could easily be a event as CD, and attaching 1 out of C-J as the third spot you could easily have CDC or CDD as an event, which is an repetition. And as I previously said repetition is not logical in this problem since these are 10 individuals and (if letter noted A-J) it is not possible to repeat the letter at any spot.