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Probability; Blind taste competition

Expert replies
by psm12se » Sun Dec 01, 2013 6:49 am
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
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Source: — Problem Solving |

by Brent@GMATPrepNow » Sun Dec 01, 2013 8:37 am
psm12se 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. 1/12
B. 5/14
C. 4/9
D. 1/2
E. 2/3
This question can be solved using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(contestant does not taste all 3 samples) = 1 - P(contestant DOES taste all 3 samples)

P(contestant DOES taste all 3 samples)
For this event to occur, the contestant must taste 2 cups of one sample, 1 cup from another sample, and 1 cup from another sample.

Let's take the task of tasting all 3 samples and break it into STAGES.
Stage 1: Select the tea that will be tasted twice. There are 3 types of tea, so stage 1 can be completed in 3 ways.

Stage 2: Choose 2 cups to taste from tea selected in stage 1. Since the order in which we select the 2 cups does not matter, we can use combinations. We can select 2 cups from 3 cups in 3C2 ways(= 3 ways).

Stage 3: From one of the two remaining (untasted) teas, select 1 cup to taste. There are 3 cups, so stage 3 can be completed in 3 ways.

Stage 4: Select 1 cup from the last remaining (untasted) tea. There are 3 cups, so stage 4 can be completed in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages in (3)(3)(3)(3) ways (= 81 ways)

The TOTAL number of ways to select 4 cups from 9 cups = 9C4 = 126

So, P(contestant DOES taste all 3 samples) = 81/126 = 9/14

This means that P(contestant does not taste all 3 samples) = 1 - 9/14
= [spoiler]5/14[/spoiler]
= B


Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Aside: If anyone is interested, we have a free video on calculating combinations (like 3C2 and 9C4) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Sun Dec 01, 2013 9:41 pm
Here's a way that gets you there very quickly:

Suppose that our three types of tea are Mint, Jasmine, and Green, and that we have three cups of each.

Now suppose that I want the probability that I don't taste any of the cups of Mint tea. My four cups can be ANY four of the other six, so there are (6 choose 4) ways that I can pick those four cups. If there were no restrictions at all, of course, there are (9 choose 4) ways that I can pick those four cups, so the odds that I try no cups of Mint tea are

(6 choose 4) / (9 choose 4)

But I need to consider the possibility of any of the three scenarios, not just Mint, so the odds are really

3 * (6 choose 4)
----------------
(9 choose 4)

which simplifies to 5/14.
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