How many sets of 6 can be made from a set of 12

[Vector1962]

Homework Statement

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A bag has a total of 12 balls, 3 of which are black. If you select 6 balls from the bag. a) How many sets of 6 can be made? b) How many of the sets of 6 contain 3 black balls? c) What is the probability of selecting a set of 6 that contains 3 black balls?

Homework Equations

_NP_R
NC_R

The Attempt at a Solution

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Part a) I believe ₁₂C₆ gives the total of sets of 6 = 924
Part b) I believe 9C6 gives the total of sets of 6 of white balls only so (₁₂C₆) - (₉C₆) = sets that have 3 black?
Part c) [ ₁₂C₆ - ₉C₆ ] / ₁₂C₆ ?

 

[FactChecker]

Part a: You haven't said if the 3 black balls are considered identical. Your calculations assume that they are all different. The same goes for the other (non-black) balls. Your answer may be right, but it may not be.

Part b: For a quick sanity check, (₁₂C₆) - (₉C₆) = 924-84=840. If that was correct, you would be almost certain (see Part c answer) to pick all 3 black balls even though they are only 1/4th of the balls. So half of the picks are black even though they are only 1/4th of the 12. That seems wrong.

Part c: (924-84)/924 = 0.90909... This seems too high since only 1/4th of the balls are black and this says you are very likely to select 3 black in 6.

 

[Dick]

Homework Statement

[/B]
A bag has a total of 12 balls, 3 of which are black. If you select 6 balls from the bag. a) How many sets of 6 can be made? b) How many of the sets of 6 contain 3 black balls? c) What is the probability of selecting a set of 6 that contains 3 black balls?

Homework Equations

_NP_R
NC_R

The Attempt at a Solution

[/B]
Part a) I believe ₁₂C₆ gives the total of sets of 6 = 924
Part b) I believe 9C6 gives the total of sets of 6 of white balls only so (₁₂C₆) - (₉C₆) = sets that have 3 black?
Part c) [ ₁₂C₆ - ₉C₆ ] / ₁₂C₆ ?

The set of all choices minus the set of choices of only white balls leaves the set of all choices containing at least one black ball. Not what you want.

 

[Vector1962]

In terms of combinations, it seems there is only 1 combination that has 3 black balls and 3 white balls?

 

[Dick]

In terms of combinations, it seems there is only 1 combination that has 3 black balls and 3 white balls?

Since you only have 3 black balls, I'd agree about there only being one choice there. But how about the white balls? You have more than 3 to choose from.

 

[Vector1962]

In terms of combinations, if I have 3 black balls and 3 white balls, that leaves 6 white balls which can be split into 2 combinations of 3 white balls. Which, in turn, can be paired with the 3 black balls. So, 3 possible sets of those... but aren't those 3 sets all the same --> 3 black and 3 white?

 

[FactChecker]

For part b: Specify a specific order of picking 3 black and 3 white balls and figure out how many ways that specific order can occur. Then multiply that by the number of ways 3 black and 3 white balls can be ordered in the 6 balls.
As always, be careful about double counting the same thing.

 

[Vector1962]

Sorry, not even a clue what you mean. If I specify BWBWBW as a specific order there is only 1 way that occurs. Are you asking me to find how many ways can 3 black balls and 3 white balls be arranged?

 

[Dick]

Sorry, not even a clue what you mean. If I specify BWBWBW as a specific order there is only 1 way that occurs. Are you asking me to find how many ways can 3 black balls and 3 white balls be arranged?

You are getting off the beam here. Of course, if you pick 3 white balls and 3 black balls you wind up with 3 black and 3 white. Number each ball and call them $B_1, B_2, B_3, W_1, W_2, W_3, W_4, W_5, W_6,W_7, W_8, W_9$. Now how many ways to choose 3 white and 3 black?

 

[FactChecker]

EDIT: CORRECTION. This post is wrong.

Sorry, not even a clue what you mean. If I specify BWBWBW as a specific order there is only 1 way that occurs. Are you asking me to find how many ways can 3 black balls and 3 white balls be arranged?

Your calculated answers like part a ₁₂C₆ have been assuming that all the balls are different regardless of their color. In other words, in your calculations B₁WB₂WB₃W is not the same as B₂WB₁WB₃W.
That is why I asked in post #2 if the black balls should be considered identical or not. If you consider them to be identical, then you have to take that into account in all your answers. Your answer in part a would be wrong. You need to divide by the number of different ways balls that you consider identical could be swapped around.

 

[Dick]

Your calculated answers like part a ₁₂C₆ have been assuming that all the balls are different regardless of their color. In other words, in your calculations B₁WB₂WB₃W is not the same as B₂WB₁WB₃W.
That is why I asked in post #2 if the black balls should be considered identical or not. If you consider them to be identical, then you have to take that into account in all your answers. Your answer in part a would be wrong. You need to divide by the number of different ways balls that you consider identical could be swapped around.

This is going off track. Combinations, like $_{12}C_6$, do NOT count different orderings as different. And I think questions a) and b) are steps on the way to answering c) which is a question about probability. If the balls are indistinguishable then you could correctly answer a) as "There are 4 ways, no black, 1 black, 2 black or 3 black". That's obvious and it's useless for calculating probability since those 4 cases aren't equally probable. I think any debate about 'identical' or not is confusing the issue.

 

[FactChecker]

This is going off track. Combinations, like $_{12}C_6$, do NOT count different orderings as different. And I think questions a) and b) are steps on the way to answering c) which is a question about probability. If the balls are indistinguishable then you could correctly answer a) as "There are 4 ways, no black, 1 black, 2 black or 3 black". That's obvious and it's useless for calculating probability since those 4 cases aren't equally probable. I think any debate about 'identical' or not is confusing the issue.

Oh! Sorry. right! I stand corrected. They do not count different orderings as different. I will correct my post.

 

[Vector1962]

Well, if I have a set of n items then the number of permutations of n items where p items are identical and q items are identical and r items are identical and so on is given by n! / (p! q! r!...) . At least that's what I found in a statistics book. So, if I have a set of 6 balls and 3 of them are black and 3 of them are white then the number of distinct permutations is 6! / (3! 3!) = 20. so if there are 20 sets possible containing 3 white balls and 3 black balls. This leads to 20 / ₁₂C₆ as the probability of getting one of those sets from the ₁₂C₆ possible sets? --> 20/924 = 0.02

 

[Vector1962]

As long as I'm guessing, The number of ways to arrange 3 black balls from a set of 3 black balls is ₃C₃ = 1. The number of ways to select 3 white balls from a set of 9 white balls ₉C₃ = 84. The fundamental counting principle indicates the number of ways to pair these 2 sets together is to multiply the number of elements in each set together, so 84 * 1 = 84. The total number of distinct sets is ₁₂C₆ = 984 hence, the probability of selecting 1 of the 84 "good sets" from the total possible sets is 84/984 = 0.08 ? --> for some reason this one sound right? Even if it's not, I appreciate the time you both devoted to help make the approach to this type of problem more clear.

 

[Dick]

As long as I'm guessing, The number of ways to arrange 3 black balls from a set of 3 black balls is ₃C₃ = 1. The number of ways to select 3 white balls from a set of 9 white balls ₉C₃ = 84. The fundamental counting principle indicates the number of ways to pair these 2 sets together is to multiply the number of elements in each set together, so 84 * 1 = 84. The total number of distinct sets is ₁₂C₆ = 984 hence, the probability of selecting 1 of the 84 "good sets" from the total possible sets is 84/984 = 0.08 ? --> for some reason this one sound right? Even if it's not, I appreciate the time you both devoted to help make the approach to this type of problem more clear.

You've got it. This version is correct. Except that $_{12}C_9=924$. It's not clear to me what the other approach is all about with counting some kind of permutations in the numerator and combinations in the denominator. If you want proof, do an experiment with real balls. This approach will give you the correct probability of drawing 3 white and 3 black.

 

[Vector1962]

It's not clear to me what the other approach is all about with counting some kind of permutations in the numerator and combinations in the denominator.

The permutation 'gibberish' is me reading a statistics book, grasping at some straws trying to make some sense of all this. In the text it mentions "repeated" items and thought I was getting close. I realize the approach should be "Read, Understand, Do". Sometimes I just have to put something down on 'paper' and see if it makes sense. To be truthful, this counting stuff get's confusing. Again, thanks for the help.

 

[PeroK]

The permutation 'gibberish' is me reading a statistics book, grasping at some straws trying to make some sense of all this. In the text it mentions "repeated" items and thought I was getting close. I realize the approach should be "Read, Understand, Do". Sometimes I just have to put something down on 'paper' and see if it makes sense. To be truthful, this counting stuff get's confusing. Again, thanks for the help.

Here's an alternative way to do part c). Maybe this will be gibberish as well!

First, if you pick 6 balls, what is the probability the first 3 are black?

That's $(\frac{3}{12})(\frac{2}{11})(\frac{1}{10}) = \frac{1}{220}$

Next, how many ways could you get all three black?

It could be the first three, numbers 1,2 and 4 etc. That's $\binom{6}{3} =20$.

Are all these possibilities equally likely? is it just as likely to get numbers 1,2,3 black as any other three numbers? You may need to think about this, but the answer is yes.

So, the probability of getting all three black is $\frac{20}{220} = \frac{1}{11}$.

This agrees with your revised answer of $\frac{84}{924}$.

Does this more intuitive approach help?

 

[Vector1962]

Does this more intuitive approach help?

I'm afraid it would have not have occurred to me to approach the problem this way. For some reason, needed to think in terms of sets and I needed a set of 6. My efforts were directed on finding how to generate sets of 6. To start by picking 3 black 'first' would not have crossed my mind. Actually, I'm not sure why I would find the probability of picking the 'first' 3 black? Your solution amounts to the probability of picking 3 black balls times the number of distinct ways 3 white and 3 black can be arranged -> n! / (p! q! r!...) ?

 

[PeroK]

I'm afraid it would have not have occurred to me to approach the problem this way. For some reason, needed to think in terms of sets and I needed a set of 6. My efforts were directed on finding how to generate sets of 6. To start by picking 3 black 'first' would not have crossed my mind. Actually, I'm not sure why I would find the probability of picking the 'first' 3 black? Your solution amounts to the probability of picking 3 black balls times the number of distinct ways 3 white and 3 black can be arranged -> n! / (p! q! r!...) ?

My solution was to realize that getting all 3 black balls can be done in 20 equally likely ways. Then you just need to calculate the probability of one of those ways and multiply by 20.

I like to do these counting problems two different ways and check I get the same answer.