And the good one again. +1 to Economist.

"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.

\(P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}\).

Answer: B.

Hope it's clear.
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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,

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

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

\(P = \frac{45}{{9*2*7}}\)

\(P = \frac{5}{14}\)

We want to select 2 cups of tea which are of similar type and the rest 2 cups which each are of different type ;
hence the terms 3C2 * 3C1* 3C1
Now you should multiply it by 3 because ; out of the 3 tea types , you may select the tea type which is in 2 cups, in 3 ways.

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Not sure if my channel of thought is flawed to start with, but when it is said that it's a 'blind' taste competition, I am considering repetitions. As the contestant does not know which of the 9 cups he/she had picked up the first time, the same can possibly be repeated in the second turn and so on until the fourth.
That gives a complete different perspective to the problem.
Where am I going wrong here?

Hi,
Let me try..
mainhoon, you are going with the logic that you select 1 cup from each sample in 3c1 x 3c1 x 3c1 ways and 1 remaining cup from 6 cups in 6c1 ways rite!
I guess the flaw with your approach is that this 6c1 is the probablity of selecting 1 item from 6 DIFFERENT items (r items from n different items is nCr)
Im this case the 6 items left are not different! They are XX,YY,ZZ types right! so you just cant use the 6c1 fomula!
This is the formula : The number of ways of choosing r objects from p objects of one kind, q objects of second kind, and so on is the coefficient of x^r in the expansion


hope it helps!

The probability of tasting all 3 cups is

One cup from pool A (\(\frac{3}{9}\)) x One cup from pool B (\(\frac{3}{8}\)) x One cup from pool C (\(\frac{3}{7}\))

Since the contestant has to taste 4 cups, one possibility is picking second cup from pool A, so the probability is \(\frac{2}{6}\)

The probability of tasting in order of ABCA is \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6}\) ..... (1)

In how many ways can the contestant taste ABCA cups = \(\frac{4!}{2!}\) = 4 x 3 ....... (2)

The combinations ABCB & ABCC are also possible (along with ABCA) = 3 combinations ..... (3)

Since the choice of "a specific cup" from "a pool" doesn't matter here, we need NOT account for \(A_1A_2A_3, A_1A_3A_2,\) ... etc. combinations

So Probability of tasting all three cups is = (1) * (2) * (3)

= \(\frac{3}{9} * \frac{3}{8} * \frac{3}{7} * \frac{2}{6} * 4 *3 *3\)

=\(\frac{9}{14}\)

Hence, Probability of NOT tasting all three cups = \(1- \frac{9}{14} = \frac{5}{14}\)
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1st cup : 3/9 is actually the probability that it's type A; Probability that it's not would be 6/9

Interesting way to approach the problem, though.

To choose 2 types of sample out of 3, we get 3c2
We then have 6 cups and to choose 4 cups out of that we get 6c4.
Hence, 3c2 * 6c4
The total ways of selecting 4 cups out of the total 9 is 9c4.
Final ans is (3c2 * 6c4) / 9c4

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

Bad outcomes:
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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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?