In the second approach why don't we multiply by \(C^1_2\) while selecting the second or third tea type?

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?

# \(\frac{1}{12}\)

# \(\frac{5}{14}\)

# \(\frac{4}{9}\)

# \(\frac{1}{2}\)

# \(\frac{2}{3}\)

And the good one again. +1 to

Economist.

"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.

\(P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}\).

Answer: B.

Hope it's clear.