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At a blind taste competition, a contestant is offered \(3\) cups of each of the \(3\) samples of tea in a random arrange

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At a blind taste competition, a contestant is offered \(3\) cups of each of the \(3\) samples of tea in a random arrange

by Vincen » Sun Oct 03, 2021 8:11 am

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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. \(\dfrac1{12}\)

B. \(\dfrac5{14}\)

C. \(\dfrac49\)

D. \(\dfrac12\)

E. \(\dfrac23\)

Answer: B

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Re: At a blind taste competition, a contestant is offered \(3\) cups of each of the \(3\) samples of tea in a random arr

by regor60 » Sun Oct 03, 2021 9:55 am
So try recognizing the various outcomes if you did pick all three, but setting aside the order in which were picked:

abca

abcb

abcc

Now, each of these can be picked in 4! ways. But each also has a duplicate tea, so need to divide by 2.

So we have 3 outcomes x 4!/2 = 36.

Now we need to account for the probability of each outcome.

There are 3 A's , so 3/9
3 B's so 3/8
3 C's so 3/7
and finally 2 A's so 2/6

3/9x3/8x3/7x2/6 = 1/3x3/8x3/7x1/3=

1/8x 1/7 = 1/56.

Each of the 36 ways has the same probability, so 36x1/56 =36/56.

But to find probability of NOT picking all 3 teas, need to subtract 36/56 from 1 = 20/56 = [spoiler]B, 5/14[/spoiler]
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