At a blind taste competition a contestant is offered 3 cups

Question

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}

Anju@Gurome · Answer

varun289 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?

Probability that a contestant does not taste all of the samples = 1 - probability that a contestant tastes all 3 samples.

This means first we should find the probability that a contestant tastes 2 cups of one sample and 1 cup from each of other 2 samples.

Total number of ways to choose 4 cups out of 9 = 9C4 = 9!/[(4!)*(5!)] = 9*8*7*6/4! = 9*2*7
Number of ways to choose the sample which provides with 2 cups = 3C1 = 3
Number of ways to choose these 2 cups from the sample chosen = 3C2 = 3
Number of ways to choose 1 cup out of 3 from the 2nd sample = 3C1 = 3
Number of ways to choose 1 cup out of 3 from the 3rd sample = 3C1 = 3

So, probability that a contestant tastes all 3 samples = (3*3*3*3)/(9*2*7) = 9/14
Therefore, required probability = 1 - 9/14 = 5/14

The correct answer is B.

Anju Agarwal
Quant Expert, Gurome

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J N · Answer

how would/or is it possible to do this with 1-prob of getting all 3

choose A, then B, then C then any

3/9 * 3/8 * 3/7 * 6/6

take this and multiple by 3! to get different abc arrangements

it does not work but wondering if can do it and how

GMATGuruNY · Answer

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

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Brent

Amrabdelnaby · Answer

Hi Guru,

can you please show me how to solve it using combinations?

Thanks

GMATGuruNY 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

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.

Brent@GMATPrepNow · Answer

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

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

Then you can try solving the following questions:

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- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html

MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
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- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
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DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
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- https://www.beatthegmat.com/please-solve ... 71499.html
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Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

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

Let's say that there are 3 kinds of tea: A, B and C, and there are 3 cups of each tea.

First find P(contestant does not taste tea A)
P(contestant does not taste tea A) = P(1st selection is not tea A AND 2st selection is not tea A AND 3rd selection is not tea A AND 4th selection is not tea A)
= P(1st selection is not tea A) x P(2st selection is not tea A) x P(3rd selection is not tea A) x P(4th selection is not tea A)
= 6/9 x 5/8 x 4/7 x 3/6
= 5/42

Now find P(contestant does not taste tea B)
The steps to find this probability will be the same as steps taken to find the above probability.
So, P(contestant does not taste tea B) = 5/42

Likewise, P(contestant does not taste tea C) = 5/42

So, P(contestant does not taste all of the samples) = 5/42 + 5/42 + 5/42
=[spoiler] 5/14 [/spoiler]
= B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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