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. \(\frac{2}{3}\)
Really good problem. Let's break it down.
-- Does order matter? No. (i.e. Are we selecting or arranging? Problem says "random arrangement," this means selecting/combinations; all 3 cups of A are the same)
-- Is there replacement? No, unless specifically stated assume there is no replacement.
-- Is each grouping of them the same? Yes, for the denominator total (fundamental counting principle) but No for the numerator (we need to find 4 cups with a certain condition). There are 2 possibilities, either 2 types of tea or 3 types of tea, we can't get 1 type of tea because there's 4 selections and only 3 cups of each type.
-- The easiest way to approach is to find P(2 teas) = 1 - P(3 teas). I made a chart using slot method:
Ways to choose first, second, third, any tea type (3 * 2 * 1 * 1)
* Number of cup choices (\(\frac{3}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))
Step by step:
3 ways to choose 1st tea type * 3 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total (anything can fulfil this requirement; we are asking for the way to choose ANY tea, rather than the way to choose one of the 3 types. Since we can choose any tea, we can choose any cup also)
After reducing fractions, we are left with 9/14. Usually GMAT problems on probability will have answers that add up to 1, so I would expect that as a trap answer. Since we found the P(3 types) and are looking for P(2 types) we need to do 1 - 9/14 = 5/14.
locklock wrote:
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?
locklock That solution is similar to mine, but the poster reasoned through it a bit differently. I think what he meant was that there's 2 distinct ways for the second/third slot to account for (Say you pick Tea A for first slot, you can get B or C for 2nd slot. If you got B, then the 3rd is A, but if you got A, then the third is B). That's why he multiplied by 2.
I would write it like this because it's more clear to me:
Ways to choose ANY tea, 2nd tea, 3rd tea, ANY tea (1 * 2 * 1 * 1) *
Number of choose cup choices ( \(\frac{9}{9} * \frac{6}{8} * \frac{3}{7} * \frac{6}{6}\))
Step by step:
1 way to choose ANY tea type * 9 cups out of 9 total
2 ways to choose 2nd tea type * 6 remaining cups out of 8 total
1 way to choose 3rd tea type * 3 remaining cups out of 7 total
1 way to choose ANY type * 6 remaining cups out of 6 total.
You can see that the difference is only where the factor of 3 goes, whether in the ways to choose or the choices, so it works out to be the same.
Does that make sense?