Permutations, Combinations and Probability: In a room there are 16 wooden chairs and 10 plastic chairs.

[yobacul]

In a room there are 16 wooden chairs and 10 plastic chairs. Except for the colour, the wooden chairs are identical and the same holds for the plastic chairs. Of the wooden chairs, 5 are red, 5 are blue and 6 are green. Of the plastic chairs, 4 are red, 2 are blue and 4 are green. In how many different ways can 9 chairs be chosen from the total number of 26 chairs in the room such that there are 3 of each colour? What is the probability that only one of the 9 chosen chairs is wooden?

Any help with this? I cannot start it off as there are so many different options for 3 of each colour - I think there must be an easier way. Thanks

 

[Dr.Peterson]

In a room there are 16 wooden chairs and 10 plastic chairs. Except for the colour, the wooden chairs are identical and the same holds for the plastic chairs. Of the wooden chairs, 5 are red, 5 are blue and 6 are green. Of the plastic chairs, 4 are red, 2 are blue and 4 are green. In how many different ways can 9 chairs be chosen from the total number of 26 chairs in the room such that there are 3 of each colour? What is the probability that only one of the 9 chosen chairs is wooden?

Any help with this? I cannot start it off as there are so many different options for 3 of each colour - I think there must be an easier way. Thanks

Never refuse to start because it looks too hard at first glance. You may learn some useful things as you work, or it may be easier than you think.

I started by putting the data into a table so it was easier to see.

Start with the numerator: how many ways are there to choose 3 red, 3 blue, and 3 green so that only one is wooden?

The denominator will be easier.

 

[yobacul]

Never refuse to start because it looks too hard at first glance. You may learn some useful things as you work, or it may be easier than you think.

I started by putting the data into a table so it was easier to see.

Start with the numerator: how many ways are there to choose 3 red, 3 blue, and 3 green so that only one is wooden?

The denominator will be easier.

Thanks for the encouragement

I thought of all 9 wooden (3R, 3B, 3G), 8 plastic (3R, 2B, 4G) and 1 wooden (1B), 2 of each wooden and 1 of each plastic, 2 of each plastic and one of each wooden ... but there are more. Like this it takes ages. I am sure there is an easier way out which I am not seeing

 

[Dr.Peterson]

Thanks for the encouragement

I thought of all 9 wooden (3R, 3B, 3G), 8 plastic (3R, 2B, 4G) and 1 wooden (1B), 2 of each wooden and 1 of each plastic, 2 of each plastic and one of each wooden ... but there are more. Like this it takes ages. I am sure there is an easier way out which I am not seeing

Please show your work in detail (at least the actual list of cases, and what you did for one case, so we can see how you are doing it). We can't tell you what is easier without knowing what it has to be easier than!

Why would you include "2 of each wooden" when we're working on "only one is wooden"?

 

[JeffM]

PKA could probably give you a formula, but a formula does not teach you how to think.

One way to structure the problem is as follows.

What are the number of ways to choose nine chairs so that all nine are of the same color? The answer is fairly obvious. There is exactly one way to choose nine red chairs from nine red chairs. There are ten ways to choose nine green chairs from ten green chairs. There is no way to choose nine blue chairs from seven blue chairs. So the answer is there are eleven ways to get nine chairs, all of the same color

Now do the same thing for eight chairs of the same color. Now for seven chairs of one color, then six, and then five.

When you get to four chairs of the same color, you have to be careful to avoid double counting. Why is that an issue? How do you address it?

Finally, compute how many ways you can select exactly three chairs of each color.

Add those seven numbers up to get the number of ways that you can select nine chairs from twenty-six.

Now what?

 

[Dr.Peterson]

I think it's simpler than JeffM says; possibly I'm interpreting something differently:

In a room there are 16 wooden chairs and 10 plastic chairs. Except for the colour, the wooden chairs are identical and the same holds for the plastic chairs. Of the wooden chairs, 5 are red, 5 are blue and 6 are green.

Of the plastic chairs, 4 are red, 2 are blue and 4 are green. In how many different ways can 9 chairs be chosen from the total number of 26 chairs in the room such that there are 3 of each colour?

What is the probability that only one of the 9 chosen chairs is wooden?

We only need to count choices of 3 of each color. We don't care about choosing 9 of one color, or anything else. The first question is not about probability.

Also, although it emphasizes the chairs of each type being identical, I think (because we are asking for probability in the second part) that we need to take them as distinct. (Imagine putting a numbered sticker on each chair.) That is another possible point of confusion; I'm hoping to see in your work where you might be confused.

 

[yobacul]

Please show your work in detail (at least the actual list of cases, and what you did for one case, so we can see how you are doing it). We can't tell you what is easier without knowing what it has to be easier than!

Why would you include "2 of each wooden" when we're working on "only one is wooden"?

The first question does not say only one is wooden.

 

[JeffM]

I think it's simpler than JeffM says; possibly I'm interpreting something differently:

I was interpreting the question as asking what is the probability that we choose 3 chairs of one color AND that one of them be wooden.

You are entirely correct that if the question is “what is the probability that of nine chairs such that the number of chairs of each color is equal,” what I suggested is way too complicated.

And if the question is “what is the probability that among nine randomly selected chairs exactly one is wooden,” again what I suggested is way too complicated.

As usual we need the EXACT wording of the problem to give a good answer. Are we dealing with joint probabilities or conditional probabilities. I was assuming the ultimate goal was a conditional probability. I may be totally wrong.

 

[JeffM]

The first question does not say only one is wooden.

If the question says something like “nine chairs are chosen at random and it turns out that there are three of each color, what then is the probability that exactly one chair is wooden,” then my previous answer was wrong.

 

[Dr.Peterson]

The first question does not say only one is wooden.

We have some misunderstandings to clear up, which are my fault.

Here is what I said, to give context:

Start with the numerator: how many ways are there to choose 3 red, 3 blue, and 3 green so that only one is wooden?

The denominator will be easier.

Since I talked about numerator and denominator, it should (in principle) be clear that I was suggesting that you think about the second part (the probability) first:

In a room there are 16 wooden chairs and 10 plastic chairs. Except for the colour, the wooden chairs are identical and the same holds for the plastic chairs. Of the wooden chairs, 5 are red, 5 are blue and 6 are green. Of the plastic chairs, 4 are red, 2 are blue and 4 are green.

In how many different ways can 9 chairs be chosen from the total number of 26 chairs in the room such that there are 3 of each colour?

What is the probability that only one of the 9 chosen chairs is wooden?

Because you mentioned wooden chairs (which are not an issue in the first part), I supposed that you were responding to what I said, first finding the numerator of the probability in the second part, which involves only one wooden chair. The denominator of that fraction is what the first part is asking for, which I was holding for last. I see that I just confused you.

So let's back up, and focus on the first question, which is not a probability but a count.

Here is what you said, presumably about that:

I thought of all 9 wooden (3R, 3B, 3G), 8 plastic (3R, 2B, 4G) and 1 wooden (1B), 2 of each wooden and 1 of each plastic, 2 of each plastic and one of each wooden ... but there are more. Like this it takes ages. I am sure there is an easier way out which I am not seeing

But the first part doesn't mention being wooden at all! We just have 9 red, 7 blue, and 10 green chairs, and need to pick 3 of each color. This is not hard, because it doesn't require paying attention to what a chair is made of.

The woodenness of a chair comes into the second part only.

So give it another try. I apologize for confusing things.

(I'm assuming, by the way, that you quoted the problem exactly.)

 

[Steven G]

I cannot start it off as there are so many different options for 3 of each colour - I think there must be an easier way. Thanks

OK, fine, there are so many different option for 3 of each color. So try using smaller numbers and learn from this new case.

 

[yobacul]

OK, fine, there are so many different option for 3 of each color. So try using smaller numbers and learn from this new case.

We have some misunderstandings to clear up, which are my fault.

Here is what I said, to give context:

Since I talked about numerator and denominator, it should (in principle) be clear that I was suggesting that you think about the second part (the probability) first:

Because you mentioned wooden chairs (which are not an issue in the first part), I supposed that you were responding to what I said, first finding the numerator of the probability in the second part, which involves only one wooden chair. The denominator of that fraction is what the first part is asking for, which I was holding for last. I see that I just confused you.

So let's back up, and focus on the first question, which is not a probability but a count.

Here is what you said, presumably about that:

But the first part doesn't mention being wooden at all! We just have 9 red, 7 blue, and 10 green chairs, and need to pick 3 of each color. This is not hard, because it doesn't require paying attention to what a chair is made of.

The woodenness of a chair comes into the second part only.

So give it another try. I apologize for confusing things.

(I'm assuming, by the way, that you quoted the problem exactly.)

 

[yobacul]

Yes, here the problem is quoted exactly. I’m finding this the hardest of the 3 questions I’ve posted as there seems to be so many options! If there’s an easier way, a tip would be appreciated. Thanks

 

[yobacul]

Yes, here the problem is quoted exactly. I thought of doing 9C3x6C3x10C3. But this seems too straightforward for 4 marks (second part carries 2 marks) and the fact it says 'different ways' I thought I had to take into consideration of all the different options - be it wood or plastic. Thanks

 

[BigBeachBanana]

Yes, here the problem is quoted exactly. I thought of doing 9C3x6C3x10C3. But this seems too straightforward for 4 marks (second part carries 2 marks) and the fact it says 'different ways' I thought I had to take into consideration of all the different options - be it wood or plastic. Thanks

There are 7 blue chairs so it should be \(\displaystyle {9 \choose 3 }{7 \choose 3 }{10 \choose 3 }\)

What is the probability that only one of the 9 chosen chairs is wooden?

Let A be the event of picking exactly 1 wooden and 8 plastic chairs. B the event of picking 9 chairs with exactly 3 colors each.
You're looking for the conditional probability [imath]\Pr(A|B) = \dfrac{|A \cap B|}{|B|}[/imath].

The first question took care of the [imath]|B|[/imath]. Now, you need to find [imath]|A \cap B|[/imath]. Try counting the elements of [imath]|A \cap B|[/imath] in three groups based on the color of the wooden chair. (Hint: some of the groups are invalid)

 

[stapel]

User "nad081" changed username to "yabacul".

Change approved.

Eliz.