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altairahmad
GMAT 1: 700 Q47 V39
Posts: 258
17 Oct 2020, 10:19
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IanStewart
Thanks for the reply.
I am unable to wrap my head around the highlighted portion. However, I took your word for it and in my original solution posted I divided the first part by 2 (i.e 36/2), keeping everything else same, and that gave me the correct answer.
I would appreciate if you could perhaps explain in different terms why the first parts counts double. I divided the permutations with number of different ways to remove order. It seemed to work well except the double counting thing which I do not understand. I understand that with FCP, both AABB and BBAA are perhaps counted, is this the reason why were are dividing by 2 ? I have been using FCP for almost all of the combinations questions. This is the first question that has given me trouble. How do I spot double counting issue ?
17 Oct 2020, 13:52
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altairahmad
As I said above, it's quite a subtle issue. It's probably easier to see if you imagine a simpler problem. Say you have two different men, and two different women, and you will select all four of them for your board of directors. You want to know the probability you get two men and two women. Obviously that has to happen, so the probability is 1 (so there's no need to do any more work, but I'll use this to illustrate why double-counting can be an issue in the tea problem). So we should get 1 if we calculate the probability using any method at all - if we don't, our method is wrong. If we imagine picking people one at a time, there will be 4! = 24 possible selections. So that will be our denominator.
Now say we decided to count the number of ways we could pick two men and two women, in order, by counting this way (I'm mirroring how you solved the tea problem) :
- we can first count how many ways we can pick, in order, XXYY, where the first two selections are the same type, and the last two are the same type
- we have 4 choices for the first selection, since it can be anything, 1 for the second (it must match the first), 2 for the third, and 1 for the last, for 8 orders in total that look like XXYY
- notice now what you're counting: how many selections you can make that look like MMWW but also how many you can make that look like WWMM, because we said X can be anything
- we haven't counted selections that look like MWMW, for example. So you might then say: we can arrange the letters X, X, Y, Y in six different orders -- I'll list them (you can just do 4!/(2!)(2!) though), and I'll also write out the possibilities each sequence encapsulates:
XXYY = MMWW or WWMM
XYXY = MWMW or WMWM
XYYX = MWWM or WMMW
YXXY = MWWM or WMMW
YXYX = MWMW or WMWM
YYXX = MMWW or WWMM
So notice we're already counting all the sequences that look like MMWW or WWMM when we count how many ways to make a sequence in the pattern XXYY. If we thought that YYXX was different from XXYY, and that the pattern YYXX gave us 8 additional sequences, we'd be counting MMWW and WWMM all over again, so we'd be counting all of those sequences exactly twice. We don't want to do that, so if we count in this way, we need to divide by 2 to account for the fact that we're double-counting. And if you do that, you get 8*6/2 = 24, which is the answer we anticipated.
The double-counting problem really arises here because we said at the outset "X can be anything". If instead we solved this way, which is much easier to think about: "the number of ways to get the sequence MMWW is (2)(1)(2)(1) = 4, and we can arrange M, M, W, W in 6 ways in total, so we have 4*6 = 24 sequences," then the issue doesn't come up at all. It's really because we introduced this extra layer of abstraction, where we decided to count "two things that match, then two other things that match" that the problem became difficult to think through correctly.
altairahmad
Three things I'll say about this:
- the only way to always avoid it is to really understand what you're counting when you solve a counting problem;
- it becomes more likely to happen though the more abstract your solution is (see the last paragraph I wrote above), since then it can become harder to see exactly what you're counting and what you're not
- this question is not at all representative of ordinary GMAT questions, so it's very unlikely you'll need to deal with a similar situation again. Double-counting can be important, though, on some realistic GMAT problems, but it's much more obvious. If you had this question:
One person attending a conference will be selected at random. What is the probability that person will be a woman or an engineer (or both)?
1. 60% of the conference attendees are women
2. 20% of the conference attendees are engineers
The answer here is E. Some people will pick C, because they will think it's fine to add the 60% and 20%. It's not though, because when you do that, you might be double-counting, when the two groups overlap. It might be the case that every engineer is a woman, and then the answer would be 0.6, but it might also be that no engineer is a woman, and then 0.8 is the correct answer (among other possibilities).
That's a simple example to illustrate that double-counting can be relevant if you have cases that might overlap. You always need to think carefully before solving a question using cases about two things: do my cases include all the situations I want them to include, and do my cases overlap? If they overlap, you usually want to start again, and define non-overlapping cases. So one last example:
Amit has one quarter (worth $0.25), two dimes (each worth $0.10) and three pennies (each worth $0.01). If he selects three of these coins at random, what is the probability the coins he selects will be worth more than $0.20?
Someone solving this problem *incorrectly* might think:
- if Amit picks a quarter, it doesn't matter what else he picks. So they might find that probability -- say it's "p" (if you solve, p = 0.5)
- if Amit picks two dimes, it doesn't matter what else he picks. So they might find that probability too -- say it's "q" (if you solve, q = 0.2)
then they might think, since there's no other way to get more than $0.20, that the answer is p+q. But it's not, because we're counting the possibility of getting one quarter and two dimes in both cases. So we're counting it twice. One correct way to solve is instead to find the probability of these two cases:
- Amit picks a quarter. The probability he picks a quarter with his first selection is 1/6, so the probability he does on any of the three selections is 3/6 = 1/2
- Amit picks two dimes and a penny (importantly, it must *not* be a quarter). The probability he picks dime, dime, penny is (2/6)(1/5)(3/4), but the penny could be picked first or second or third, for three orders in which we get two dimes and one penny, so the answer is (3)(2/6)(1/5)(3/4) = 3/20
Now we have accounted for all possible cases, and have not double-counted anything, so we can just add to get the answer.
If you recognize that double-counting is an issue in these simpler examples, you'll very likely be fine for the GMAT.
04 Jul 2022, 15:36
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Economist
I bet you can't find an OG P/C question that's as tough as this one, but it's a fun one and I didn't see a response above that covers this method, so here goes.
Rather than doing the 1-minus technique, I went straight for the probability of NOT tasting all three.
Let's make the teas A, B, and C. Note that I use X and Y below. That is because I figure out how to select SOME tea as X and a DIFFERENT tea as Y. Which is which (and which is left out) is irrelevant.
How can we get a tasting that doesn't have one of each of A, B, and C? There are only two possibilities.
Some arrangement of XXXY, where X is one type of tea and Y is some other type.
Or
Some arrangement of XXYY, where X is one type of tea and Y is some other type.
Let's look at XXXY first.
What's the probability of getting exactly that arrangement? (9/9)*(2/8)*(1/7)*(6/6) = 1/28
There are 4 ways to arrange XXXY (XXXY, XXYX, XYXX, YXXX...yes, this is 4C1).
4*(1/28) = 4/28
Let's look at XXYY.
What's the probability of getting exactly that arrangement? (9/9)*(2/8)*(6/7)*(2/6) = 1/28
There are 6 ways to arrange XXYY (XXYY, XYXY, XYYX, YXXY, YXYX, YYXX...yes, this is 4C2).
6*(1/28) = 6/28
(4/28)+(6/28) = 10/28 = 5/14
Answer choice B.
Edited to add:
The fractions that we multiply to find, say, XYXX are obviously not exactly the same as those that we multiple to find XXXY, but the numerators are exactly the same (we have, in some order, 9*2*1*6) and the denominators are exactly the same (9*8*7*6). That's why we can just multiply by the 4 arrangements of XXXY rather than calculating the probability of each arrangement separately. Each arrangement of Xs and Y just rearranges the numerators, but the value of the fraction stays the same.
08 May 2023, 23:41
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I used a different method.
Assume the cups of tea to be A1,A2,A3,B1,B2,B3,C1,C2,C3.
4 cups out of 9 cups can be selected using 9C4.
For a contestant to not taste all 3 types of tea, the contestant can taste 2 types of tea in the following ways.
2 cups of one type + 2 cups of another type
3C2 * 3C2 = 9
There are total 3 such cases possible (type AB, BC, AC).
Hence 9 * 3 = 27
3 cups of one type + 1 cup of another type
3C3 * 6C1 = 6
Again there are total 3 such cases possible (type AB, BC, AC)
6 * 3 = 18
Therefore total cases possible = 18 + 27 =45
Probability = 18/126 = 5/14
Option B
Assume the cups of tea to be A1,A2,A3,B1,B2,B3,C1,C2,C3.
4 cups out of 9 cups can be selected using 9C4.
For a contestant to not taste all 3 types of tea, the contestant can taste 2 types of tea in the following ways.
2 cups of one type + 2 cups of another type
3C2 * 3C2 = 9
There are total 3 such cases possible (type AB, BC, AC).
Hence 9 * 3 = 27
3 cups of one type + 1 cup of another type
3C3 * 6C1 = 6
Again there are total 3 such cases possible (type AB, BC, AC)
6 * 3 = 18
Therefore total cases possible = 18 + 27 =45
Probability = 18/126 = 5/14
Option B
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