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Combinatorics: Taking out balls with different colors

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Combinatorics: Taking out balls with different colors  [#permalink]

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New post 14 Dec 2018, 15:46
Hello guys! I am a bit confused about a theoretical issue on probabilities and combinatorics. I'll use an example to make things clearer:

Let's assume we have a bag with 5 red balls, 4 green balls and 3 blue balls.

1) If the question is: in how many different ways can we take all the balls out of the bag? The answer would be: (12!)/(5!*4!*3!).

2) On the other hand, if the question was: what is the probability of taking 2 red balls and 1 green ball in three random picks? The answer would be: (5/12)*(4/11)*(4/10)*(3!/2!).

3) However (and now it's when I'm getting really confused!), what would be the answer if the question was: in how many different ways can we take 3 balls out of the bag? According to the second question it should be 12*11*10. But, according to the first question there is a risk of overcalculation since we can take balls that are of the same color, which means their order doesn't matter. Can someone please explain the rationale behind it and clarify what would be the correct answer for the question 3?

Sorry for the long post, but it was the only way that I found to make myself clear.

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Re: Combinatorics: Taking out balls with different colors  [#permalink]

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New post 21 Dec 2018, 16:45
pedroabreu wrote:
Hello guys! I am a bit confused about a theoretical issue on probabilities and combinatorics. I'll use an example to make things clearer:

Let's assume we have a bag with 5 red balls, 4 green balls and 3 blue balls.

1) If the question is: in how many different ways can we take all the balls out of the bag? The answer would be: (12!)/(5!*4!*3!).


That depends on whether you care about the order or not.

If you want to know how many different orders you can take the balls out of the bag in, your calculation is right!

If you don't care about the orders, the answer is actually just '1' - there's only one way to take all of the balls out of the bag, if you don't care what order they come out in. :)

Quote:
2) On the other hand, if the question was: what is the probability of taking 2 red balls and 1 green ball in three random picks? The answer would be: (5/12)*(4/11)*(4/10)*(3!/2!).


Here, your answer is the 'order doesn't matter' one. If you just want two red balls and one green ball, and you don't care what order they show up in, that's the calculation you'd do.

Another way to get the same result is to look at it like this: how many different ways can you pull two red balls out of a bag with 5 red balls? That's 5*4 / 2. How many different ways can you pull one green ball out of a bag with 4 green balls? That's just 4. So, there are 5*4*4 / 2 = 80/2 = 40 ways to pull out two red balls and one green ball. Since we're doing probability, we need a denominator: the number of ways you could pull any three balls out of a bag containing 12 balls. That's 12*11*10 / (3*2*1) = 220.

The probability is 40/220, or 4/22, or 2/11. That's the same as you get from your calculation, if you simplify it.

The reason they're different is because we took basically the same steps in a slightly different order. You started by finding the probability of drawing the balls in a certain order: red, then red, then green. Then, you multiplied that by the number of different orders, since the probability of each of the 3 possible orders would be the same. In other words, you started with the 'order matters' probability, and then multiplied it by a factor to change it to 'order doesn't matter'.

In my approach above, I started with 'order doesn't matter', and just focused on finding the number of ways we could pull the balls out, regardless of what order they came out in. Because I didn't start with order, I didn't have to multiply by 3 (or 3!/2!) at the end.

Both approaches give you the same answer, and I don't think either one is better.

This might be why you're starting to get confused - your answer to the first question is the 'order matters' answer, and your answer to the second question is the 'order doesn't matter' answer. That's why it seems like they're giving you contradictory information when you try to answer the third question.


Quote:
3) However (and now it's when I'm getting really confused!), what would be the answer if the question was: in how many different ways can we take 3 balls out of the bag? According to the second question it should be 12*11*10. But, according to the first question there is a risk of overcalculation since we can take balls that are of the same color, which means their order doesn't matter. Can someone please explain the rationale behind it and clarify what would be the correct answer for the question 3?

Sorry for the long post, but it was the only way that I found to make myself clear.


This is actually a really complex question, and not one that you'd see on the test. But let's solve it anyways to satisfy intellectual curiosity!

Let's decide first whether we want order to matter. What does that mean here, anyways? If order matters, that just means that RED RED BLUE is different from RED BLUE RED, which is different from BLUE RED RED, etc. But if order doesn't matter, then we want to count all three of those as the same. They would have to make that clear in the problem. But because this is a theoretical question, not an official GMAT problem, we can decide whether we think order should matter or not :)

Let's say that order does matter. So, we want to consider RED BLUE GREEN to be different from GREEN BLUE RED, to be different from GREEN RED BLUE, etc.

In that case, the number of ways to draw three balls out of the bag is just 3*3*3 = 27. That seems weird, but you can double-check it by listing out all of the possibilities:

1 R R R
2 G G G
3 B B B

4 R R B
5 R B R
6 B R R
7 R B B
8 B R B
9 B B R
10 R R G
11 R G R
12 G R R
13 R G G
14 G R G
15 G G R
16 B B G
17 B G B
18 G B B
19 B G G
20 G B G
21 G G B

22 R G B
23 R B G
24 G B R
25 G R B
26 B R G
27 B G R

Now, that doesn't mean that the probability of each of these results is the same. For instance, we're more likely to draw R R R than G G G, because there are more red balls in the bag than there are green balls.

We also got to 'cheat' a little bit, since there are at least three balls of each color in the bag. If there were only two blue balls, for example, we'd have to cross off the B B B case.

You're probably more interested in the probability of each of those possibilities, though. For instance, what is the probability of drawing R G R? Since order matters, we'd just have to multiply the three probabilities together: 5/12 * 4/11 * 4/10 = 80/1320 = 2/33.

Or, what is the probability of drawing R R R? That's 5/12*4/11*3/10 = 60/1320 = 1/22.

And what if order doesn't matter? How many possibilities are there then? In this case, it's probably easier just to list them:

R R R
G G G
B B B

R R G
R G G
R R B
R B B
B G G
B B G

R B G

There are 10 different possibilities. Again, not all 10 have the same probability, which is where things get interesting. Suppose you wanted the probability of drawing one red, one blue, and one green in any order. In that case, you could either take the probability as calculated previously, and multiply it by 3!. Or, you could take 5 * 4 * 3 / (12 * 11 * 10). Same result, either way.

Finally: why do neither of these results look like your original approach to this question? I believe that's because you got mixed up with whether the balls are distinct from each other. The 12*11*10... etc. would be part of your calculation only if all 12 of the balls could be distinguished from each other. That's a possibility! But, based on your earlier two questions, I assumed that we can't tell the red balls apart from each other, can't tell the green balls apart from each other, etc.
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Re: Combinatorics: Taking out balls with different colors   [#permalink] 21 Dec 2018, 16:45
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