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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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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I like the solution - it’s helpful.

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14 May 2026, 13:07

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Can we not use the approach where not select all = 1 - selecting all three type of cup please someone explain this. i have just done 1 - 3C1*3C1*3C1*1/9C4 (Ans has not came in this way)

Bunuel

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14 May 2026, 20:27

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sagarkacha

Can we not use the approach where not select all = 1 - selecting all three type of cup please someone explain this. i have just done 1 - 3C1*3C1*3C1*1/9C4 (Ans has not came in this way)

Yes, we can use the complement approach, but your counting of “selecting all three types” is incomplete.

Since 4 cups are selected and there are only 3 tea types, tasting all 3 types must mean the distribution is 2, 1, 1.

Your expression 3C1 * 3C1 * 3C1 counts only one cup from each type, which accounts for only 3 cups, not 4.

We must also choose which tea type gives 2 cups:

3C1 * 3C2 * 3C1 * 3C1 / 9C4 = 9/14

So the required probability is:

1 - 9/14 = 5/14.

This is explained in the official solution HERE. Please review it!

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14 May 2026, 22:39

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So lets suppose i have select 1 cup from each then after i will have 6 option available so for last selection so there should be 6C1.

then it should be 3C1*3C1*3C1*6C1/9C4..? what is wrong with this approach kindly help with this.

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15 May 2026, 01:49

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sagarkacha

That approach is not right because it overcounts the same set of 4 cups.

After choosing 1 cup from each type, the “extra” 4th cup could have been chosen as the extra cup in more than one way.

For example, A1, A2, B1, C1 is counted once when A1 is chosen first and A2 is extra, and again when A2 is chosen first and A1 is extra.

So the method counts duplicates.

Here is more on this question: https://gmatclub.com/forum/at-a-blind-t ... 86830.html Hope it helps.