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I hope here we have assumed 3 cups as different( Sample 1 in one cup is different from sample 1 from second cup) What if 3 cups are identical that means no particular distinction and has 3 different samples
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I think here is the easiest approach for anyone to nail this question.

We have 3 groups of teas, let's call them Group A, B, C. As we need to find the probability of tasting only 2 groups in 4 picks, then in each picking streak we are ignoring one group.

So here is the multiplication:

First picking streak First pick: 6/9 (because 1 group is not participating - let's say we overlook group C, we have 6 choices out of 9) Second pick: 5/8 (we keep ignoring group C, and we have 5 choices left as 1 has been picked) 3rd pick: 4/7 4th pick: 3/6 6/9 x 5/8 x 4/7 x 3/6 = Probability of picking just 2 groups in 4 picks ignoring group C. We need to add another picking streaks as we have 2 groups left so you just multiple all that probability by 3.

6/9 x 5/8 x 4/7 x 3/6 x 3 = 5/14

If you think this is helpful, please give KUDOS. Thank you.

Number of ways to pick 2 types of cups : 3C2 = 3. Number of ways to pick 4 cups from these two types already chosen : 3C1*3C3 + 3C2*3C2 + 3C3*3C1 = 15. Number of ways to pick 4 cups without the said restriction = 9C4 = 9*2*7.

Hence, required probability = (15*3)/9C4 = (15*3)/(9*2*7) = 5/14.

I think this is a poor-quality question and I don't agree with the explanation. Not to taste all of the samples means anything less than 3 which could be 2, 1 or 0

Re: D01-38 : What am I missing here in my approach
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14 Sep 2018, 09:41

Hi, What am I missing in my approach ?

I have the following the following solutions.

I partition the set into samples of flavors A, B, and C. And I count the event that a candidate has not tasted all of the three as follows.

A B C Case I : 3 1 0 :: Probability that candidate has tasted flavor A 3 times and flavor B one time, but not flavor C. Case II: 3 0 1 Case III: 0 1 3 Case IV: 0 3 1 Case V : 1 0 3 Case VI: 1 3 0

For case I, the way I compute the probabilities are somewhat like balls in an urn. Questions I asked myself ones. Is the sample space condensed since I am leaving out one flavor each time. So, instead of 9 would it be 6 ? If so: Case I:( (3/6)*(2/5)*(1/4)*(3/3) ) Case II - VI: Same.

So the total probability for this (3,1,0) type is : 6 * ( (3/6)*(2/5)*(1/4)*(3/3) ) = 6* (1/20) = 3/10

Case VII: 2 0 2 Case VIII:0 2 2 Case IX: 2 2 0

Probability for Case VII: (3/6)*(2/5)*(3/4)*(2/3) Cases VIII and IX: Same

So the total probability for this (2,2,0) type is : 3 * ((3/6)*(2/5)*(3/4)*(2/3)) = 3*(2/20) = 3/10

Finally, The probability that be added: 3/10 + 3/10 = 6/10

Now that's not the answer. Well, am I missing a case or TYPE ? OR AM I NOT DIVIDING BY 9 ?

Assume that I am dividing my now by 9, instead of 6.

Then CASE I--VI : ( (3/9)*(2/8)*(1/7)*(3/6) ) *6 = 1/28

Then CASE VII--IX: ((3/9)*(2/8)*(3/7)*(2/6))*3 = 1/28

Dear Experts, Pls help where my approach is wrong.

Choosing only 2 samples: 2 cups of 1 sample and 2 cups of another sample: 3C2*3C2*3C2 (choose any 2 sample out of 3*choose 2 cups out of 3 for the 1st sample*choose 2 cups out of 3 for the 2nd sample): 3*3*3=27 3 cups of 1 sample and 1 cup of another sample: 3C2*3C3*3C1(choose any 2 sample out of 3*choose 3 cups out of 3 for the 1st sample*choose 1 cup out of 3 for the 2nd sample): 3*1*3=9 Total ways of choosing 4 sample out of 9: 9C4: 126 Total: 27+9/126: 2/7.

Please let me know wherr am i going wrong. Thnks & Regads