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D01-38

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New post 16 Sep 2014, 00:13
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At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

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

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New post 16 Sep 2014, 00:13
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Official Solution:

At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

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


"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 \(\gt\)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 (\(2 samples*3 \text{cups each} = 6 cups\));

\(C^4_9\) - total # of ways to choose 4 cups out of 9.

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 choose these 2 cups from the chosen sample;

\(C^1_3\) - # of ways to choose 1 cup out of 3 from second sample;

\(C^1_3\) - # of ways to choose 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
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New post 20 Nov 2014, 21:45
Bunuel wrote:
Official Solution:

\(C^4_9\) - total # of ways to choose 4 cups out of 9.


Hi Bunuel,

I'm having trouble understanding the format you are using in your explanation. In the combinatorics thread in your signature, it says:

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: \(C^n_k\)

So in your explanation, this would mean choosing 9 objects out of only 4 distinct objects. How is this possible? We would be getting a fraction here when we should be getting whole numbers.
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New post 21 Nov 2014, 04:52
OutOfTheHills wrote:
Bunuel wrote:
Official Solution:

\(C^4_9\) - total # of ways to choose 4 cups out of 9.


Hi Bunuel,

I'm having trouble understanding the format you are using in your explanation. In the combinatorics thread in your signature, it says:

A combination is an unordered collection of k objects taken from a set of n distinct objects. The number of ways how we can choose k objects out of n distinct objects is denoted as: \(C^n_k\)

So in your explanation, this would mean choosing 9 objects out of only 4 distinct objects. How is this possible? We would be getting a fraction here when we should be getting whole numbers.


\(C^4_9\), \(C^9_4\), 9C4 are the same: choosing 4 out of 9. How can we choose 9 out of 4? Those are just different ways of writing the same.
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New post 12 Dec 2014, 06:54
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New post 12 Jan 2015, 08:52
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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
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New post 30 May 2016, 14:43
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At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

So, we have :
Sample of tea A : the cups are A1, A2, and A3
Sample of tea B : the cups are B1, B2, and B3
Sample of tea C : the cups are C1, C2, and C3

So the contestant is offered : 4 random cups from the set {A1, A2, A3, B1, B2, B3, C1, C2, C3}

We want to calculate the probability to not choose the 3 samples with 4 cups ; for instance A1A2A3B2, or B13BC1C2, or C2A1A2A3, or A1A2B2B3, etc.
->So the contestant could test ONLY 2 of the 3 samples

Proba = \(\frac{(Number.of.ways.to.choose.4.cups.from.2.samples)}{(Number.total.of.ways.to.choose.4.cups.from.3.samples)}\)


1/ Number of ways to choose 4 cups from 2 samples :

Stage 1 : Choose 2 samples from the 3 = \(\frac{3*2}{2}\) = 3
We obtain : A, A, A, B, B, B for example

Stage 2 : Choose 4 from the 6 cups selected = \(\frac{6*5*4*3}{4*3*2}\) = 15

Counting Principle : 3*15 ways

2/ Number of ways to choose 4 cups from 3 samples = from 9 cups :

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


So, proba = \(\frac{3*15}{9*7*2}\) = \(\frac{5}{14}\)
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New post 30 Jun 2016, 08:23
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My "layman" approach:


At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

Simply calculating \(\frac{#possible.cases}{#tot. cases}\) we can find the solution quite fast.

First possible cases: \(9*5*4*3\)
We have 9 different ways in which we can taste the first cup of tea, after this first one we we do 9-1 (last draw) -3 one family of teas, so the following draws are 5, 4 and 3.

Total cases: \(9*8*7*6\)
Nothing surprising here.

\(\frac{(9*5*4*3)}{(9*8*7*6)}\) = \(\frac{5}{28}\) we still have to multiply * 2 because there are 2 ways in which the 2nd the 3rd cup can be drawn -> AB and BA.

So \(\frac{5}{28} * 2\) = \(\frac{5}{14}\)
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New post 12 Jul 2016, 12:01
I think this is a high-quality question.
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New post 25 Sep 2016, 01:35
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IMO this is a great question.

Here is my approach:

Given: 3 Samples of tea and for each sample we have 3 cups.

Samples of tea : Cups of tea
A : cup1, cup2, cup3
B : cup4, cup5, cup6
C : cup7, cup8, cup9

Contestant tastes 4 different cups of tea ==> It means he will taste any 4 cups from the 9 cups.

We are asked to find the probability that "a contestant does not taste all of the samples".

"a contestant does not taste all of the samples"--> means contestant have to taste any 2 samples out of the 3 (A,B,C).

Contestant can taste samples as i) A or B ii) B or C iii) A or C

How many ways contestant can does so => Select 4 cups out of 6 cups (\(6C4\)) ==> 15 ways
It can be done in 3 ways (AB, BC, or AC)

==> Total number of ways a contestant does not taste all of the samples = 3 * 15 = 45

Total ways to select 4 cups out of 9 cups = \(9C4\) = 126

Proability = 45/126 = 5/14
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New post 25 Feb 2017, 12:00
Here is a simple solution:
P(does not taste all samples) = 1 - P(does taste all samples)
P(does taste all samples) means he tastes A, B, C, and 1 more (e.g. ABCA, ABCB, ABCC)
Find the probability of one happening
P(ABCA) = 3/9 * 3/8 * 3/7 * 2/6 = 1/56
Then multiply by the # of ways to arrange each scenario: 4!/2! = 12 (mississippi rule)
Then multiply by 3 to account for all three scenarios
(1/56)*(12)*(3) = 9/14
Then subtract from 1 to get the answer:
1-(9/14) = 5/14

Answer B
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New post 06 Apr 2017, 10:56
lets consider the case of the contestant tasting all the samples ;
Cases : AABC ,ABBC , ABCC
For each case the selection is 3c2x3c1x3c1 : so, total is 3X3c2x3c1x3c1
Now the probability of contestant to taste all the three samples is (3X3c2x3c1x3c1)/(9c4) = 9/14
the probability that a contestant does not taste all of the samples : 1-9/14 = 5/14.
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New post 17 Apr 2017, 11:52
This alternative solution of filling spaces method also helps:

We have 9 cups in total, 3 cups per each of the 3 samples.

It is an selection question (order does not matter, irrelevant way we choose the samples)

Number of ways of choosing 4 cups of all the samples:
(9 x 8 x 6 x 3) / (1 x 2 x 3 x 4) =54=9 x 6

Number of ways of choosing 4 cups
(9 x 8 x 7 x 6) / (1 x 2 x 3 x 4) =9 x 14

Probability of not tasting all samples = 1 - probability tasting all samples

= 1- (9x6 / 9x14) = 5/14
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New post 04 Jun 2017, 07:58
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cledgard wrote:
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


cledgard

Would not be the probability of not choosing 1st cup Type A = 6/9? (where 6=3 of B + 3 of C, and 9= total # of cups)
2nd, not Type A = 5/8
3rd, not Type A = 4/7
4th, not Type A = 1/2 since 3/6
and 3 options to start with A, B, or C

so we would get \(\frac{6}{9}*\frac{5}{8}*\frac{4}{7}*
\frac{1}{2}*3=\frac{5}{14}\)
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New post 21 Jan 2018, 12:21
1-probability of tasting all types = probability of not tasting all types
ways to have tasted all types = AABC, BBAC, CCAB i.e 3. also note we can rearrange each in 4!/2! ways. therefore total number of options = 3x4x3
now calculate the probability of AABC = 3/9 x 2/8 x 3/7 x 3/6
note from above that this particular event i.e. AABC can occur 3x4x3 ways therefore multiply this with the probability calculated for AABC = 9/14
therefore, subtract 9/14 from 1 = 5/14
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New post 20 May 2018, 08:43
I think this is a high-quality question and I agree with explanation.
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New post 22 Jul 2018, 10:04
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I think here is the easiest approach for anyone to nail this question.

We have 3 groups of teas, let's call them Group A, B, C.
As we need to find the probability of tasting only 2 groups in 4 picks, then in each picking streak we are ignoring one group.

So here is the multiplication:

First picking streak
First pick: 6/9 (because 1 group is not participating - let's say we overlook group C, we have 6 choices out of 9)
Second pick: 5/8 (we keep ignoring group C, and we have 5 choices left as 1 has been picked)
3rd pick: 4/7
4th pick: 3/6
6/9 x 5/8 x 4/7 x 3/6 = Probability of picking just 2 groups in 4 picks ignoring group C. We need to add another picking streaks as we have 2 groups left so you just multiple all that probability by 3.

6/9 x 5/8 x 4/7 x 3/6 x 3 = 5/14

If you think this is helpful, please give KUDOS. Thank you.
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New post 06 Feb 2019, 22:42
Bunuel

Regarding the other way to solve this problem: Calculating the opposite probability and then substracting the same from 1,

Here is my logic:
First we select one cup from each of the r different varieties: 3C1 X 3C1 X 3C1
For the fourth cup we can choose one from any of the remaining 6 cups : 6C1 ways.
Total no of events is 9C4

So the reqd probability for the opp event stands at (3C1 X 3C1 X 3C1)/9C4....
But this is >1...
Please explain where am i going wrong
Thanks
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New post 05 Apr 2019, 23:26
One simple solution from my side:
Suppose the cups are like A,A,A,B,B,B,C,C,C => 3A, 3B, 3C
Not selecting all the 3 types of samples means selecting only 2 types of samples.

Now, possible ways of doing so is either 2A + 2B or 3A + 1B OR 1A + 3B.

multuply this by 3 as we can select A &B or A & C or B & C.

Solving this, we get [u]3C1 X 3C2 + 3C3 X 3C1 + 3C1 X 3C3/9C4 = 5/14
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New post 21 Apr 2019, 10:50
I think this is a high-quality question and I agree with explanation.

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Re D01-38   [#permalink] 21 Apr 2019, 10:50

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