D01-38

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