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# At a blind taste competition a contestant is offered 3 cups of each of

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Director
Joined: 04 Aug 2010
Posts: 536
Schools: Dartmouth College
At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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22 Feb 2019, 12:20
Economist 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. $$\frac{1}{12}$$

B. $$\frac{5}{14}$$

C. $$\frac{4}{9}$$

D. $$\frac{1}{2}$$

E. $$\frac{2}{3}$$

P(good outcome) = 1 - P(bad outcome)

A bad outcome occurs when all 3 flavors are sampled.
For all 3 flavors to be sampled, 1 of the 3 flavors must be chosen TWICE.
Number of options for the twice-chosen flavor = 3. (Any of the 3 flavors.)
From 3 cups of the twice-chosen flavor, the number of ways to choose 2 cups = 3C2 = (3*2)/(2*1) = 3.
From 3 cups of the next flavor, the number of ways to choose 1 cup = 3.
From 3 cups of the last flavor, the number of ways to choose 1 cup = 3.
To combine these options, we multiply:
3*3*3*3

All possible outcomes:
From 9 cups, the number of ways to choose 4 = 9C4 = $$\frac{9*8*7*6}{4*3*2*1} = 9*2*7$$

Thus:
P(bad outcome) = $$\frac{bad-outcomes}{all-possible-outcomes} = \frac{3*3*3*3}{9*2*7} = \frac{9}{14}$$
P(good outcome) = $$1 - \frac{9}{14} = \frac{5}{14}$$

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Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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28 Jul 2019, 11:09
AKProdigy87 wrote:
I got B: 5/14 as well, although using a different approach.

Number of ways to choose 4 cups to drink from: 9C4

Since the contestant must drink 4 cups of tea, and there are only 3 cups of each tea, the contestant MUST drink at least two different samples.

Number of ways to choose which two samples the contestant drinks (since he can't drink all 3): 3C2

Three cases:

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3

Therefore,

$$P = \frac{3C2(3C3*3C1 + 3C2*3C2 + 3C1*3C3)}{9C4}$$

$$P = \frac{3(1*3 + 3*3 + 3*1)}{\frac{9*8*7*6}{4*3*2*1}}$$

$$P = \frac{45}126$$

$$P = \frac{5}{14}$$

this is how i tried to do in a way, however, want to know between a and c, if i multiple by 3 to consider any flavor, why do i need another combination of it.

a) 3 Cups of Sample 1, 1 Cup of Sample 2: 3C3 * 3C1 * 3 for any flavor
b) 2 Cups of Sample 1, 2 Cups of Sample 2: 3C2 * 3C2 *3 for any flavor
c) 1 Cup of Sample 1, 3 Cups of Sample 3: 3C1 * 3C3 -> Why do i need combination C
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Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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16 Sep 2019, 07:38
Economist 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. $$\frac{1}{12}$$

B. $$\frac{5}{14}$$

C. $$\frac{4}{9}$$

D. $$\frac{1}{2}$$

E. $$\frac{2}{3}$$

Breakdown: There are 3 cups with three different samples and there are total of 9 cups .
So all 9 cups we can say ,
Sample 1--contained in 3 similar cups.
Sample 2--contained in 3 similar cups
Sample 2--contained in 3 similar cups

Total number of ways in which one can taste all the three samples(from a set of 4)
= 3C1(sample 1,one cup) * 3C1(sample 2,1cup)* 3C2(sample 3,2 cup) *3(total ways in which these three selections can be made(3!/2)=3)
=3*3*3*3= 81 ways

Required probability = 1 - 81/9c4 = 5/14
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Posts: 82
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WE: Business Development (Computer Software)
Re: At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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03 Nov 2019, 06:47
locklock wrote:
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = $$\frac{9}{9}$$
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = $$\frac{6}{8}$$
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = $$\frac{3}{7}$$
the fourth cup can be selected from any of the remaining 6 cups. P = $$\frac{6}{6}$$

there are $$2$$ ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = $$\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}$$

required probability = $$1 - \frac{9}{14} = \frac{5}{14}$$

I am having issues with the above, especially this line
there are $$2$$ ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?

please explain why we multiplied by 2 I cannot relate it with other examples such as https://gmatclub.com/forum/as-part-of-a ... 56446.html
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Joined: 11 Jul 2019
Posts: 1
At a blind taste competition a contestant is offered 3 cups of each of  [#permalink]

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07 Nov 2019, 20:27
locklock wrote:
dimitri92 wrote:

i solved the question as follows. i hope my approach is correct.

let's consider that the contestant tasted all 3 flavors.
the first cup can be selected from any 9 cups. P = $$\frac{9}{9}$$
the second cup can be selected from 6 (other 2 flavors) of the remaining 8 cups. P = $$\frac{6}{8}$$
the third cup can be selected from 3 (flavor not tested yet) of the remaining 7 cups. P = $$\frac{3}{7}$$
the fourth cup can be selected from any of the remaining 6 cups. P = $$\frac{6}{6}$$

there are $$2$$ ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

probability that the contestant tasted all flavors = $$\frac{9}{9}*\frac{6}{8}*\frac{3}{7}*\frac{6}{6}*2 = \frac{9}{14}$$

required probability = $$1 - \frac{9}{14} = \frac{5}{14}$$

I am having issues with the above, especially this line
there are $$2$$ ways in which second & third cup can be tested {AB, BA}. first & fourth cup already have probability 1.

in the 2nd fraction, 6/8, you already considered that either the 2nd or 3rd group can be considered. Why still a need for x2?

Bunuel, Gladiator59, generis - I have been racking my head over why that 2 is added. Please help us gain clarity.
At a blind taste competition a contestant is offered 3 cups of each of   [#permalink] 07 Nov 2019, 20:27

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