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At a blind taste competition a contestant is offered 3 cups of each of
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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?
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14 Nov 2009, 08:06
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Economist wrote:
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?
"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.
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06 Feb 2011, 08:18
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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 = \(1 - \frac{9}{14} = \frac{5}{14}\)
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15 Nov 2009, 04:16
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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
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15 Aug 2010, 18:43
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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.
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15 Aug 2010, 19:03
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We want to select 2 cups of tea which are of similar type and the rest 2 cups which each are of different type ; hence the terms 3C2 * 3C1* 3C1 Now you should multiply it by 3 because ; out of the 3 tea types , you may select the tea type which is in 2 cups, in 3 ways.
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15 Aug 2010, 20:06
Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this?
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05 Mar 2013, 09:23
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?
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06 Mar 2013, 02:32
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tosattam wrote:
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.
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09 Mar 2013, 22:37
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Hi, Let me try.. mainhoon, you are going with the logic that you select 1 cup from each sample in 3c1 x 3c1 x 3c1 ways and 1 remaining cup from 6 cups in 6c1 ways rite! I guess the flaw with your approach is that this 6c1 is the probablity of selecting 1 item from 6 DIFFERENT items (r items from n different items is nCr) Im this case the 6 items left are not different! They are XX,YY,ZZ types right! so you just cant use the 6c1 fomula! This is the formula : The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of x^r in the expansion
hope it helps!
mainhoon wrote:
Yes I can see that clearly I am ending up with more combinations than necessary. I was thinking if there is way to factor out those repeats the way I am doing it. I see your point in the simple example. However when I think of this, I am thinking of a 3x3 matrix where I have to make sure I get one from each row and the 4th can come from anywhere - that is the same thing as saying 2 from a row and 1 each from the remaining 2 (your and correct approach). My count differs from yours by a factor of 2 (I am 162 and you are 81). How can I rationalize this?
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12 Jan 2015, 08:47
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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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Re: At a blind taste competition a contestant is offered 3 cups of each of
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07 Apr 2016, 07:09
Is this an okay way to think about this problem?
In the case where we want to use the alternate solution (1-opposite), would it be okay to say the following:
The number of ways to taste all the cups would result from the following choices:
(A A) B C
- Of the 3 samples, choose 1 to be a double - Of the 3 samples, choose 2 to be singles - Given our choice for the double, choose 1 location of 3 possible to represent the double - Given our choice for the singles, choose 2 locations of the 3 possible to represent the singles
Hence our total desired outcome would be: (3C1)(3C2)(3C1)(3C2) / (9C4) = 9/14. Our answer, therefore, would be 1 - 9/14 = 5/14
Re: At a blind taste competition a contestant is offered 3 cups of each of
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06 May 2017, 10:22
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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
1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9
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25 Sep 2017, 11:15
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DanceWithFire wrote:
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
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
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Re: At a blind taste competition a contestant is offered 3 cups of each of
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15 Feb 2018, 05:47
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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
At a blind taste competition a contestant is offered 3 cups of each of
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22 Feb 2019, 13:20
Economist wrote:
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. \(\frac{2}{3}\)
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 = \(\frac{9*8*7*6}{4*3*2*1} = 9*2*7\)
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At a blind taste competition a contestant is offered 3 cups of each of
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