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. 1/12
B. 5/14
C. 4/9
D. 1/2
E. 2/3
P(NOT selecting all 3 types of tea) = 1 - P(selecting all 3 types of tea).
Let the 3 types of tea be A, B and C.
For all 3 types of tea to be tasted, one type must be selected EXACTLY TWICE.
P(exactly n times) = P(one way) * (total possible ways).
One way to select A exactly twice: AABC
P(A is selected 1st) = 3/9. (Of the 9 cups, 3 contain A.)
P(A is selected 2nd) = 2/8. (of the 8 remaining cups, 2 contain A.)
P(B is selected 3rd) = 3/7. (Of the 7 remaining cups, 3 contain B.)
P(C is selected 4th) = 3/6. (Of the 6 remaining cups, 3 contain C.)
Since we want all of these events to happen, we multiply the probabilities:
3/9 * 2/8 * 3/7 * 3/6 = 1/56.
Total possible ways to select A exactly twice:
AABC is only ONE WAY for A to be selected exactly twice.
Since arrangement of AABC represents a different way to select A exactly twice, the result above must be multiplied by the number of ways to arrange AABC.
The number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways to arrange the identical elements.
The reason: when the identical elements swap positions, the arrangement doesn't change.
Since AABC includes two identical elements, we divide by 2!:
Number of ways to arrange AABC = 4!/2! = 12.
Thus:
Number of ways to select A exactly twice = 1/56 * 12 = 3/14.
BBAC and CCAB:
Since any of the 3 teas could be the one selected exactly twice, the result above must be multiplied by 3:
Thus:
P(selecting all 3 types of tea) = 3/14 * 3 = 9/14.
Thus:
P(not selecting all 3 types of tea) = 1 - 9/14 = 5/14.
The correct answer is
B.
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