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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

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

Source: GMAT Club Tests

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Source: Beat The GMAT — Problem Solving | Sun Oct 03, 2021 8:11 am

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

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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