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Bunuel
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D01-38 22 Feb 2023, 07:32
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I updated the wording of both the question and the solutions. Hope now it's better!
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D01-38 08 May 2023, 23:37
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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
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D01-38 26 Jun 2023, 03:34
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Bunuel
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. \(\frac{2}{3}\)
Show more


Hi Bunuel,
I aaproached this way:
1-P(tastes all three types of tea)

Fav Ways= 3C1.3C1.3C1.6C1
Total Ways= 9.8.7.6

What am i doing wrong? Please help
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D01-38 26 Jun 2023, 04:02
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Bunuel
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. \(\frac{2}{3}\)


Hi Bunuel,
I aaproached this way:
1-P(tastes all three types of tea)

Fav Ways= 3C1.3C1.3C1.6C1
Total Ways= 9.8.7.6

What am i doing wrong? Please help
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I hope the following post here might help you.
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D01-38 12 Aug 2023, 10:41
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I think this is a high-quality question and I agree with explanation.
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D01-38 03 Jan 2024, 10:27
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Bunuel
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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very elegant solution- thanks!

Quick Q though- in your second solution - I tried doing it this way:

1 - 3 (number of ways of choosing tea A) *3 (number of ways of choosing tea B)*3 (number of ways of choosing tea C)*2(number of ways of choosing tea A)


Would you know what's the flaw in my approach please?
Thanks!
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D01-38 03 Jan 2024, 12:00
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tickledpink001
Bunuel
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

very elegant solution- thanks!

Quick Q though- in your second solution - I tried doing it this way:

1 - 3 (number of ways of choosing tea A) *3 (number of ways of choosing tea B)*3 (number of ways of choosing tea C)*2(number of ways of choosing tea A)


Would you know what's the flaw in my approach please?
Thanks!
Show more

In your calculation, you've only considered one specific pattern of selection, like AABC. However, this method overlooks other possible combinations such as BBAC, or CCAB. Additionally, when calculating for a pattern like AABC, the correct method isn't to multiply 3*3*3*2, but rather 3C2*3*3. This accounts for choosing 2 out of 3 cups from one type and one each from the others. Therefore, the total should be calculated as 3(3C2*3*3), representing all possible arrangements.

Hope it's clear.
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D01-38 04 Jan 2024, 06:48
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Bunuel
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Bunuel
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

very elegant solution- thanks!

Quick Q though- in your second solution - I tried doing it this way:

1 - 3 (number of ways of choosing tea A) *3 (number of ways of choosing tea B)*3 (number of ways of choosing tea C)*2(number of ways of choosing tea A)


Would you know what's the flaw in my approach please?
Thanks!

In your calculation, you've only considered one specific pattern of selection, like AABC. However, this method overlooks other possible combinations such as BBAC, or CCAB. Additionally, when calculating for a pattern like AABC, the correct method isn't to multiply 3*3*3*2, but rather 3C2*3*3. This accounts for choosing 2 out of 3 cups from one type and one each from the others. Therefore, the total should be calculated as 3(3C2*3*3), representing all possible arrangements.

Hope it's clear.
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clear, thank you!!
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D01-38 07 Apr 2024, 03:19
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Hi Bunuel why is arrangement not considered here?
Bunuel
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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­
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D01-38 07 Apr 2024, 03:40
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unicornilove
Hi Bunuel why is arrangement not considered here?
Bunuel
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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­In both approaches, the order is disregarded both in the numerator and the denominator, effectively balancing each other out.
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For the negation/opposite approach, why is it not - (3C1 * 3C1 * 3C1 * 6C1)/ 9C4? We select 1 cup from each type by 3*3C1, after that there are 6 cups remaining, and any one of these 6 can be selected at random for the fourth cup.
Bunuel
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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D01-38 20 Dec 2024, 11:28
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For the negation/opposite approach, why is it not - (3C1 * 3C1 * 3C1 * 6C1)/ 9C4? We select 1 cup from each type by 3*3C1, after that there are 6 cups remaining, and any one of these 6 can be selected at random for the fourth cup.
Bunuel
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 hope the following post here might help you.
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D01-38 08 Feb 2025, 22:25
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I like the solution - it’s helpful.
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I did not quite understand the solution. How was choosing 4 cups out of 9 as 9C4...because there are repeating type of teas(3 of a kind)
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This is because all 9 cups are different when nothing of similarity is mentioned in question. The tea inside cup may be same but here choosing the cup will be from selecting 4 out of 9 cups without repetition.
ayushman0301
I did not quite understand the solution. How was choosing 4 cups out of 9 as 9C4...because there are repeating type of teas(3 of a kind)
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D01-38 25 Jul 2025, 02:33
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I like the solution - it’s helpful.
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theRishabhSh
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D01-38 06 Aug 2025, 06:06
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I did not quite understand the solution. Why we are not considering the order in which the person picks the cups and instead solving through combinations?
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Bunuel
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D01-38 06 Aug 2025, 10:33
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theRishabhSh
I did not quite understand the solution. Why we are not considering the order in which the person picks the cups and instead solving through combinations?
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When using the combination approach, we can disregard order. However, if you check the discussion, you'll find several alternative solutions using the probability approach, where order is considered.
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D01-38 11 Aug 2025, 02:00
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This is a great question that’s helpful for learning and I like the solution - it’s helpful. Unfortunately even though I selected 5/14 the question status is showing that i got it wrong.
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D01-38 13 Sep 2025, 06:01
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Hi, I understand your solution. My only question is why are we not taking into account the ways in which these 4 cups can be arranged - so
1. multiply the denominator by 4!
2. multiply the numerator by 4!/2! (since either of the 3 cups are repeated in one of the cases)
Bunuel
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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