And the good one again. +1 to Economist.

"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).

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

\(C^2_3\) - # of ways to choose which 2 samples will be tasted;
\(C^4_6\) - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups);
\(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

Another way:

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

Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).

\(C^1_3\) - # of ways to choose the sample which will provide with 2 cups;
\(C^2_3\) - # of ways to chose these 2 cups from the chosen sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample;
\(C^1_3\) - # of ways to chose 1 cup out of 3 from third sample;
\(C^4_9\) - total # 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.

Hope it's clear.
_________________

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.

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,

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

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

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

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

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.

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?

A blind taste simply means that a contestant doesn't know which samples he/she is offered to taste.
_________________

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 = 3/9
2nd cup, Probability it is not type A = 3/8, since we only have 8 cups left, and 3 of them are type A
3rd cup, Probability it is not type A = 3/7
4th cup, Probability it is not type A = 3/6
We do the same for types B and C (multiply by 3)
So : 3/9 * 3/8 * 3/7 * 3/6 * 3 = 5/14
_________________

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

1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9

Interesting way to approach the problem, though.

You are right, I made a mistake when I wrote it no the forum. The correct approach is the following:
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
_________________

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

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


_________________

I am having issues with the above, especially this line
there are \(2\) ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?

Really good problem. Let's break it down.

-- Does order matter? No. (i.e. Are we selecting or arranging? Problem says "random arrangement," this means selecting/combinations; all 3 cups of A are the same)
-- Is there replacement? No, unless specifically stated assume there is no replacement.
-- Is each grouping of them the same? Yes, for the denominator total (fundamental counting principle) but No for the numerator (we need to find 4 cups with a certain condition). There are 2 possibilities, either 2 types of tea or 3 types of tea, we can't get 1 type of tea because there's 4 selections and only 3 cups of each type.
-- The easiest way to approach is to find P(2 teas) = 1 - P(3 teas). I made a chart using slot method:

Ways to choose first, second, third, any tea type (3 * 2 * 1 * 1) * Number of cup choices (\(\frac{3}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))

Step by step:
3 ways to choose 1st tea type * 3 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total (anything can fulfil this requirement; we are asking for the way to choose ANY tea, rather than the way to choose one of the 3 types. Since we can choose any tea, we can choose any cup also)

After reducing fractions, we are left with 9/14. Usually GMAT problems on probability will have answers that add up to 1, so I would expect that as a trap answer. Since we found the P(3 types) and are looking for P(2 types) we need to do 1 - 9/14 = 5/14.




locklock That solution is similar to mine, but the poster reasoned through it a bit differently. I think what he meant was that there's 2 distinct ways for the second/third slot to account for (Say you pick Tea A for first slot, you can get B or C for 2nd slot. If you got B, then the 3rd is A, but if you got A, then the third is B). That's why he multiplied by 2.

I would write it like this because it's more clear to me:

Ways to choose ANY tea, 2nd tea, 3rd tea, ANY tea (1 * 2 * 1 * 1) * Number of choose cup choices ( \(\frac{9}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))

Step by step:
1 way to choose ANY tea type * 9 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total.

You can see that the difference is only where the factor of 3 goes, whether in the ways to choose or the choices, so it works out to be the same.

Does that make sense?

this is how i tried to do in a way, however, want to know between a and c, if i multiple by 3 to consider any flavor, why do i need another combination of it.

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1 * 3 for any flavor
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2 *3 for any flavor
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3 -> Why do i need combination C

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

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.
_________________

VeritasKarishma AjiteshArun MentorTutoring chetan2u


As per question,we want that all samples not be tasted.







No of ways :-3C1*2C1

No of favorable Outcome=3C1*3C1*3C1*3C1*2C1=162

What is wrong with my thought process?
_________________

By the method you have used above or even the member who has posted above your post, you are going wrong as you are counting each case twice, so you have to divide your all methods by 2.

How??

Say you have \(A_1,A_2,A_3\) and similarly \(C_1,C_2,C_3\) and \(B_1,B_2,B_3\).
You choose 1 of each that is \(A_1+B_1+C_1\) and 4th could any of remaining, say A2, so \((A_1+B_1+C_1)+A_2\)
But next you choose \(A_2+B_1+C_1\) and 4th could any of remaining, that is it can be A1, so \((A_2+B_1+C_1)+A_1\)

So you can see although you have counted it twice, but the elements are same.
That is why \(\frac{162}{2}=81\)

Therefore, choose 2 of 1 and 1 from the remaining two => \(3C2*3C1*3C1\), and these 2 can be any of the three colors, so \(3C2*3C1*3C1*3\)
Total = \(9C4=\frac{9*8*7*6}{4*3*2}=9*14\)

Let the probability we are looking for be p(x). Probability of n(x)=\(\frac{3C2*3C1*3C1*3}{9*14}=\frac{9}{14}\) ..
=> \(p(x) = 1-n(x) = 1-\frac{9}{14}=\frac{5}{14}\)
_________________