At a blind taste competition a contestant is offered 3 cups of each of

Official Solution:

In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?

A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)


To ensure that the contestant does not taste all three types of tea, he must taste only two types of tea. Notice that it's impossible for him to taste just one type of tea, since he will be tasting four cups and in any case, will have to try more than one type of tea.

The number of ways to taste only two types of tea is:

\(\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}\):

\(C^2_3\) is the number of ways to choose which 2 types of tea out of 3 will be tasted;

\(C^4_6\) is the number of ways to choose 4 cups out of 6 cups from these 2 types of tea (2 types = 6 cups);

\(C^4_9\) is the total number of ways to choose 4 cups out of 9.

Another way:

Calculate the probability of opposite event and subtract this value from 1.

The opposite event will be for the contestant to taste ALL 3 samples, so the contestant must choose 2 cups from one type of tea, and 1 cup from each of the other two types (2-1-1).

\(C^1_3\) is the number of ways to choose the type of tea which will provide with 2 cups;

\(C^2_3\) is the number of ways to choose these 2 cups from the chosen type;

\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the second type;

\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the third type;

\(C^4_9\) is the total number of ways to choose 4 cups out of 9.

\(P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}\).


Answer: B
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I got B: 5/14 as well, although using a different approach.

Number of ways to choose 4 cups to drink from: 9C4

Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.

Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2

Three cases:

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3

Therefore,

\(P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}\)

\(P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}\)

\(P = \frac{45}{{9*2*7}}\)

\(P = \frac{5}{14}\)

I solved it in a way that I think is more direct:

I get the probability that each cup is not a specific type of tea (type A):
1st cup, Probability it is not type A = 6/9
2nd cup, Probability it is not type A = 5/8, since we only have 8 cups left, and 3 of them are type A
3rd cup, Probability it is not type A =4/7
4th cup, Probability it is not type A = 3/6
We do the same for types B and C (multiply by 3)
So : 6/9 * 5/8 * 4/7 * 3/6 * 3 = 5/14
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The probability of tasting all 3 cups is

One cup from pool A (\(\frac{3}{9}\)) x One cup from pool B (\(\frac{3}{8}\)) x One cup from pool C (\(\frac{3}{7}\))

Since the contestant has to taste 4 cups, one possibility is picking second cup from pool A, so the probability is \(\frac{2}{6}\)

The probability of tasting in order of ABCA is \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6}\) ..... (1)

In how many ways can the contestant taste ABCA cups = \(\frac{4!}{2!}\) = 4 x 3 ....... (2)

The combinations ABCB & ABCC are also possible (along with ABCA) = 3 combinations ..... (3)

Since the choice of "a specific cup" from "a pool" doesn't matter here, we need NOT account for \(A_1A_2A_3, A_1A_3A_2,\) ... etc. combinations

So Probability of tasting all three cups is = (1) * (2) * (3)

= \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6} * 4 *3 *3\)

=\(\frac{9}{14}\)

Hence, Probability of NOT tasting all three cups = \(1- \frac{9}{14} = \frac{5}{14}\)

It should be 1-(3*3C2*3C1*3C1/9C4)
= 1-9/14 = 5/14

Now why is your approach wrong ?
E.g: in how many ways 2 objects can be selected from a set of 3.
We know it is 3C2
As per the approach you took above it should be equal to; 3C1*2C1 != 3C2.

Hope it helps.

That solution doesn't even appear to be correct, so it's unfortunate that it's been kudo'ed so many times that it appears at the top of the thread. It seems it's multiplying by 2 somewhere to massage an incorrect answer into a correct one, and because we're multiplying by 2s and 3s no matter how we solve the problem, doing that can produce a right answer even when a method is wrong.

The starting assumption behind this solution is flawed. The solution assumes the first three cups of tea are of three different types. But for a person to taste all three types of tea, the first three cups do not need to be different. Perhaps the first three are Darjeeling, Darjeeling and Orange Pekoe. Then as long as the fourth cup is Oolong, the contestant will taste all three types (I don't drink tea, so I hope these are actual types of teas!). The calculation turns out to be right, but the rationale doesn't appear to be (or if it is, I'm not understanding the explanation).

It's possible to solve the problem this way, by finding the probability of tasting all three types and subtracting from 1, though it's probably my least favourite method to do this question (mira93's solution above is the way I'd prefer). If the three types of tea are A, B and C, we can start by finding the probability of tasting the teas in this precise order: A, A, B, C. That probability is (3/9)(2/8)(3/7)(3/6). But there are 4!/2! different sequences in which we can arrange the four letters A, A, B, C (we're just counting "words" with repeated letters), so the probability the contestant tastes A twice, and the other two teas once, in some order, is (4!/2!)(3/9)(2/8)(3/7)(3/6). Because the contestant might taste B or C twice instead of A, we need to multiply this by 3 to find the probability they taste all three types of tea. So the probability the contestant tastes every type of tea in the first four cups is

(3)(4!/2!)(3/9)(2/8)(3/7)(3/6) = (3)(4)(3)(1/3)(1/4)(3/7)(1/2) = 9/14

and 1 - 9/14 = 5/14 is the answer.

P(good outcome) = 1 - P(bad outcome)

Bad outcomes:
A bad outcome occurs when all 3 flavors are sampled.
For all 3 flavors to be sampled, 1 of the 3 flavors must be chosen TWICE.
Number of options for the twice-chosen flavor = 3. (Any of the 3 flavors.)
From 3 cups of the twice-chosen flavor, the number of ways to choose 2 cups = 3C2 = (3*2)/(2*1) = 3.
From 3 cups of the next flavor, the number of ways to choose 1 cup = 3.
From 3 cups of the last flavor, the number of ways to choose 1 cup = 3.
To combine these options, we multiply:
3*3*3*3

All possible outcomes:
From 9 cups, the number of ways to choose 4 = 9C4 = \(\frac{9*8*7*6}{4*3*2*1} = 9*2*7\)

Thus:
P(bad outcome) = \(\frac{bad-outcomes}{all-possible-outcomes} = \frac{3*3*3*3}{9*2*7} = \frac{9}{14}\)
P(good outcome) = \(1 - \frac{9}{14} = \frac{5}{14}\)


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To choose 2 types of sample out of 3, we get 3c2
We then have 6 cups and to choose 4 cups out of that we get 6c4.
Hence, 3c2 * 6c4
The total ways of selecting 4 cups out of the total 9 is 9c4.
Final ans is (3c2 * 6c4) / 9c4

Posted from my mobile device

Posted from my mobile device

A blind taste simply means that a contestant doesn't know which samples he/she is offered to taste.
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There are two separate issues with this solution. You first count how many ways to get the specific sequence AABB. But to get two teas each of two types, we could have various orders: ABAB, say, or BAAB. As you correctly calculate, there are 4!/(2!*2!) orders we can put these four letters in. So to find all of the sequences of A, A, B, B, you need to multiply by 4!/(2!)(2!), and not divide by it.

But then there is a second issue, which is quite subtle, that only affects the analysis of the first case. If you assume the first tea can be anything, the second must match, and then the final teas must match each other, you'll be counting the possibilities of getting AABB and also of getting BBAA. But then if you count all the orders you can put A, A, B, B in, you'll be rearranging AABB to get BBAA, and you'll thus be counting BBAA for a second time. So doing this double-counts every possibility, and we need to divide by 2 to get the right answer. That double-counting issue is the reason many solutions earlier in this thread proceeded by first selecting the two types of tea (3C2, or 3 ways). Then you get the sequence AABB in 3C2*(3)(2)(3)(2) ways, with no double-counting.

Finally, if you're solving this way, you're assuming you're selecting the four teas in order, so the denominator needs to be 9*8*7*6, and not 9C4. Putting all of that together, the answer becomes:

[ ((9)(2)(6)(2)/2) * 4!/(2!*2!) + (9)(2)(1)(6) * 4 ] / (9)(8)(7)(6) = (9*2*6*6 + 9*2*6*4)/(9)(8)(7)(6) = (12 + 8)/8*7 = 20/56 = 5/14

I think it's tricky to get to the right answer using this precise method, so I'd definitely prefer a different approach here.



There are a couple of issues here. First, there are many ways to select one cup from each type of tea. Maybe the first two cups both are the same type of tea, and the final two are different. Or maybe the first cup comes from the third type of tea, not the first. You haven't accounted for any of those possibilities. There are solutions earlier in this thread that correctly count how many ways to pick one cup of each type, which you may want to look at. Just as importantly, when you're counting the ways to get one cup of each type, you're counting as if you're selecting cups in order. That's perfectly fine, but if you do that in the numerator of a probability, you also need to count the total number of possibilities (the denominator) as if you're selecting in order. So if you use this approach, your denominator would need to be 9*8*7*6, and not 9C4.



I don't understand any of this solution once it gets to the third cup (it frankly doesn't make any sense at all to me -- if your first two cups are AA, the third cup can literally be anything). I think it's just coincidence that you've arrived at the right answer. Conceptually, the instant a solution assumes the first two teas are both "A", it cannot possibly be correct. That's because when the first two teas are the same (both A), it becomes less likely you'll try all three types of tea (compared to the situation when the first two teas are different). Since the probability you will get all three types of tea is different when you pick AA than when you pick AB/AC, it becomes mandatory to divide the problem into cases at that point.

You can adjust that solution. If we first pick any tea at all, call it 'A', then:

• 2/8 of the time we pick A on the second selection, and 1/7 of the time we pick A with our third selection. Then the fourth selection doesn't matter, so we get a sequence like AAAX (2/8)(1/7) of the time
• 2/8 of the time we pick A on the second selection, and 6/7 of the time we pick something else with our third selection. Then the fourth selection must either be 'A' or must match the third, so there are 3 things we can pick out of 6 possibilities, and the probability we get a sequence starting with AAX (where X is not A) is (2/8)(6/7)(3/6), if we don't want all three types of tea
• 6/8 of the time we pick something other than A on the second selection. Then our third and fourth selections must either be A, or must match our second selection. So we have 4 options with our third selection, and 3 for our last selection, and we get a sequence starting AX (where X is not A) that does not have all three types of tea (6/8)(4/7)(3/6) of the time

Those are all the possible cases, and adding them, we get

(2/8)(1/7) + (2/8)(6/7)(3/6) + (6/8)(4/7)(3/6) = (1/4)(1/7 + 3/7) + (3/4)(4/7)(1/2) = 1/7 + 3/14 = 5/14

But, as I said in an earlier post, the method I think is best for this problem is the one mira93 used: it's easy to work out the probability you don't get tea A on any selection (that's just (6/9)(5/8)(4/7)(3/6)), and then we multiply by 3 to get the answer, because we want to count the possibilities where we avoid tea B and tea C also.

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.



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.

* This post has been edited to rectify an error.

If you tried to approach this question by working up to the answer, counting valid cups—e.g., 9/9 * 8/8 * 4/7 * 3/6—you would still get 4/14, and perhaps being a hair away from (B) would give you some confidence that you were on the right track. I would hope you would look at 1/12 (A) and 4/9 (C) or greater and think of them as less probable, in any case.

- Andrew
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It doesn't and that is why you need to multiply by 4!/2!.

The 9 cups are all distinct - say C1 to C3 have Tea1, C4 to C6 have Tea2 and C7 to C9 have Tea3.

Now if you were select 3 cups, you would do it in 9C3 ways. Then you would arrange them in 3! ways if order mattered.

Note that in our original question when we write (3/9 * 3/8 * 3/7 * 1), we are not just selecting. We are picking a cup with Tea1 (3/9), a cup with Tea2 (3/8) and a cup with Tea3 (3/7) and then any cup (say another Tea2).

So this is the probability of picking Tea1, Tea2, Tea3 and Tea4 in that order only.
But hey, the order does not matter. So Tea2, Tea3, Tea1, Tea2 should be acceptable too and same for all other arrangements of Tea1, Tea2, Tea3 and Tea2. That is why you multiply by 4!/2! (because order DOES NOT matter). So you need to include all arrangements in which one can pick 3 distinct teas. Every such order is acceptable.

Correct.



When you choose 3C1 and then 2C1, you are doing it from the same bunch so you are double counting. You need to directly select 2 of the 3. In your case, say you select T1 first and then T3. You take this as one case. Next time, you select T3 first and then T1. You take this an another case. But they both are the same because there is no arrangement.

So this is what you do:
You choose 2 samples out of the 3 samples which will be tasted in 3C2 ways.
For each of the 2 samples, you select 2 of the 3 cups in 3C2 * 3C2 ways.
You get 3C2 * 3C2 * 3C2 = 27 ways


Total = (18 + 27) / 9C4 = 5/14

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = \(\frac{9}{9}\)
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = \(\frac{6}{8}\)
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = \(\frac{3}{7}\)
the fourth cup can be selected from any of the remaining 6 cups. P = \(\frac{6}{6}\)

there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = \(\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}\)

required probability = \(1 - \frac{9}{14} = \frac{5}{14}\)

!	This solution is not correct. Check here as to why.

Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth.
That gives a complete different perspective to the problem.
Where am I going wrong here?

Dear IanStewart

A lot of students have trouble with the highlighted line above.
I'm a bit not sure why we need to multiply by 2! instead of 4! (since we select 4 cups of tea) ?

Please share your thought on this sir

Why does the order in which it is tasted matter over here? VeritasKarishma chetan2u

My approach: \(1- (\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*\frac{6}{6}*\frac{4!}{2!})\)

Question is why is 4!/2! necessary. We're multiplying the probabilities. It doesn't matter in whichever order he tastes, right?

Hi VeritasKarishma could you explain for me why my solution is wrong?
My approach: 2 ways for the contestant can't taste all 3 samples: He tastes 3 cups of 1 samples and 1 cup of another sample OR he tastes 2 cups of each 2 samples.
- 1st way: choose 1 sample out of 3 to provide 3 cups: 3C1. Then, 1 sample out of the remaining 2 to provide 1 cup => 2C1 * 3C1
=> 3C1 * 2C1 * 3C1 = 18
- 2nd way: choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2. Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> 3C1 * 3C2 * 2C1 * 3C2 = 54
=> total probability = (18+54)/ 9C4 = 4/7

I can't understand what's wrong with my approach. Please help. Thanks so much.