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GMAT Diagnostic Test Question 38 Field: probability Difficulty: 750

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?

In order to answer the question we need to find the overall number of outcomes and the number of favourable outcomes. The favourable outcome in this case is the one when a contestant tastes only 2 kinds of tea out of 3 kinds available. If there are 3 cups of every kind of tea, the number of favourable outcomes is calculated in the following way:

\(C_6^4 * 3 = 3 * \frac{6!}{4!*2!} = 45\)

We had to multiply by 3 because there are 3 ways the two kinds of tea could be selected from 3 available kinds.

In order to answer the question we need to find the overall number of outcomes and the number of favourable outcomes. The favourable outcome in this case is the one when a contestant tastes only 2 kinds of tea out of 3 kinds available. If there are 3 cups of every kind of tea, the number of favourable outcomes is calculated in the following way:

\(C_6^4 * 3 = 3 * \frac{6!}{4!*2!} = 45\)

We had to multiply by 3 because there are 3 ways the two kinds of tea could be selected from 3 available kinds. \(P = \frac{45}{126} = \frac{5}{14}\)

I don't really understand how you got 45. I think you might be double counting since you are basically choosing 2 out of 6. However, that still leaves the possibility that the 3rd choice might make a complete set of 3... can you clarify why you are doing this again?

Basically, after computing the total number of combinations, I counted the number of "winning" combinations. If you are choosing 4 out of 9 from 3 sets of 3 and you don't want there to be a complete set, the only possibly combinations are either 3-1-0 or 2-2-0. If you choose 3 of 1 sample, you can either choose 3-0-1 or 3-1-0. Since there are 3 samples in all, that makes 3*2 = 6 combinations.

If choose 2 of one sample and 2 of another, then the only combinations are 2-2-0, 2-0-2, and 0-2-2.

The total number of combinations is then 9, so the probability is 9/126 = 1/14.

Can someone tell me if/where I went wrong?

Also, the question needs to be more clear. You need to state that its 9 marked cups of 3 samples each, otherwise you have to assume that the samples are equally divided.

Hello bipolarbear, you are basically missing combinations:

3 0 1 x 3 = 3 3 1 0 x 3 = 3 0 1 3 x 3 = 3 1 0 3 x 3 = 3 0 3 1 x 3 = 3 1 3 0 x 3 = 3 (I multiply by 3 because in the sample with 1 cup you have 3 alternatives)

2 0 2 x 9 = 9 0 2 2 x 9 = 9 2 2 0 x 9 = 9 (you have 3 different 2-cup combination in each 2-cup sample, so 3x3)

TOTAL \(\frac{45}{126} = \frac{5}{14}\) _________________

Also, the question needs to be more clear. You need to state that its 9 marked cups of 3 samples each, otherwise you have to assume that the samples are equally divided.

Thank you! Very good point - revising right now. _________________

GMAT Diagnostic Test Question 39 Field: probability Difficulty: 750

Rating:

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. \(\frac{2}{3}\)

The highlighted part is not very clear however it is a basic combination problem.

1. 3 types of tea each has 3 cups totaling 9 cups. 2. Select 4 cups of tea of 2 different types out of 9 cups.

The number of ways 2 types of teas can be selected from 3 types of tea = 3c2 = 3 ways The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways The number of ways 4 cups of tea can be selected from 2 types of tea in 3 different ways = 3x15 = 45 ways The number of ways 4 cups of tea can be selected from all 3 types of tea = 9c4 = 9x8x7x6x5!/(5!4!) = 146 ways

The prob that 4 cups of tea can be selected only from 2 types of tea = 45/126 = 15/42 ways

I do not see any complication with the OA and OE. _________________

The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways

Hi, can someone pls explain how this is 6c2, as there is no way to distinguish between the cups of the same sample. The post of gmatclub math book (math-combinatorics-87345.html) clearly says that:

"Number of ways to pick 1 or more objects from n identical objects = n"

In this case, there are 3 ways a given cup can be picked up from one set of 3 cups and there are 2 ways to pick up that set of cups itself..i.e. there are 2 * 3 = 6 ways of picking up 4 cups.

Hence 3c2 * 6/9c4 = 1/7 should be the answer..? Pls explain. Eagerly waiting.

The number of ways 4 cups of tea can be selected from only 2 types of tea = 6c2 = 15 ways

Hi, can someone pls explain how this is 6c2, as there is no way to distinguish between the cups of the same sample. The post of gmatclub math book (math-combinatorics-87345.html) clearly says that:

"Number of ways to pick 1 or more objects from n identical objects = n"

In this case, there are 3 ways a given cup can be picked up from one set of 3 cups and there are 2 ways to pick up that set of cups itself..i.e. there are 2 * 3 = 6 ways of picking up 4 cups.

Hence 3c2 * 6/9c4 = 1/7 should be the answer..? Pls explain. Eagerly waiting.

Thanks

This question was posted in PS subforum as well, so below is my solutions from there. Hope these solutions will help to clear your doubts.

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?

"The probability that a contestant does not taste all of the samples" means that contestant tastes only 2 samples of tea (one sample is not possible as contestant tastes 4 cups>3 of each kind).

\(\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}\).

\(C^2_3\) - # of ways to choose which 2 samples will be tasted; \(C^4_6\) - # of ways to choose 4 cups out of 6 cups of two samples (2 samples*3 cups each = 6 cups); \(C^4_9\) - total # of ways to choose 4 cups out of 9.

Answer: B.

Another way:

Calculate the probability of opposite event and subtract this value from 1.

Opposite event is that contestant will taste ALL 3 samples, so contestant should taste 2 cups of one sample and 1 cup from each of 2 other samples (2-1-1).

\(C^1_3\) - # of ways to choose the sample which will provide with 2 cups; \(C^2_3\) - # of ways to chose these 2 cups from the chosen sample; \(C^1_3\) - # of ways to chose 1 cup out of 3 from second sample; \(C^1_3\) - # of ways to chose 1 cup out of 3 from third sample; \(C^4_9\) - total # of ways to choose 4 cups out of 9.

To calculate the number of favourable outcomes let us consider two cases . 1.three cups of a single sample and one cup from from the other two samples . Number of ways =3*6c1=18.[ 3-->denotes number of ways in which three cups from a single sample can be selected ]

2.two cups each from any two samples . Number of ways =3*3c2*3c2=27.[3-->denotes the number of ways of selecting 2 samples out of the 3 samples ]

Firstly, we have to eliminate the probability of taking all the samples after 4 tastes. There are 3 ways to do that.

1/ He took 3 different samples in the 3 first tastes, so we don't care whatever the last time he took. Probability of this way = 1*6/8*3/7*1=9/28

2/ The 2nd time he took the same sample with the 1st one and the 3rd and 4th time he took the 2 others. Probability = 1*2/8*6/7*3/6=3/28

3/ He took 2 different samples in the 2 first time; the 3rd time he took the same sample with either the 1st or the 2nd one, and the 4th time he took the other sample. Probability = 1*6/8*4/7*3/6=6/28