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?
A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
P(good outcome) = 1 - P(bad outcome)
Bad outcomes:A bad outcome occurs when all 3 flavors are sampled.
For all 3 flavors to be sampled, 1 of the 3 flavors must be chosen TWICE.
Number of options for the twice-chosen flavor = 3. (Any of the 3 flavors.)
From 3 cups of the twice-chosen flavor, the number of ways to choose 2 cups = 3C2 = (3*2)/(2*1) = 3.
From 3 cups of the next flavor, the number of ways to choose 1 cup = 3.
From 3 cups of the last flavor, the number of ways to choose 1 cup = 3.
To combine these options, we multiply:
3*3*3*3
All possible outcomes:From 9 cups, the number of ways to choose 4 = 9C4 = \(\frac{9*8*7*6}{4*3*2*1} = 9*2*7\)
Thus:
P(bad outcome) = \(\frac{bad-outcomes}{all-possible-outcomes} = \frac{3*3*3*3}{9*2*7} = \frac{9}{14}\)
P(good outcome) = \(1 - \frac{9}{14} = \frac{5}{14}\)
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