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}

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Bunuel

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13 Sep 2025, 06:09

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amritabehura

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)

Both the denominator and numerator are written in terms of unarranged combinations, so they balance each other out. For more, you can check the previous discussion.

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27 Oct 2025, 05:41

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This is a great question that’s helpful for learning.

singhssudh

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01 Nov 2025, 00:55

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I did not quite understand the solution. Are these 3 cups for same type, different or similar?

Bunuel

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01 Nov 2025, 01:29

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singhssudh

I did not quite understand the solution. Are these 3 cups for same type, different or similar?

Each set of 3 cups belongs to the same type of tea, so all 3 cups of Type A are the same tea, all 3 of Type B are the same, and all 3 of Type C are the same. Please check the previous three pages of discussion for more context.

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01 Nov 2025, 11:04

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singhssudh

I did not quite understand the solution. Are these 3 cups for same type, different or similar?

The question is pretty clear: ... 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. ...

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10 Dec 2025, 11:36

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I like the solution - it’s helpful.

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16 Dec 2025, 17:32

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I like the solution - it’s helpful.

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16 Feb 2026, 22:46

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I did not quite understand the solution. Why is it incorrect to think 3C1 * 3C1 * 3C1 * 6C1 (to reflect, choosing 1 cup of the 3 cups from Type A, one from Type B, and another from Type C, then with the remaining 6 cups, you pick one (hence 6C1)? Thank you.