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At a blind taste competition a contestant is offered 3 cups of each of
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VeritasKarishma
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Re: At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  10 Aug 2020, 02:11
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harshbirajdar wrote:
Why does the order in which it is tasted matter over here? VeritasKarishma chetan2u

My approach: \(1- (\frac{3}{9}*\frac{3}{8}*\frac{3}{7}*\frac{6}{6}*\frac{4!}{2!})\)

Question is why is 4!/2! necessary. We're multiplying the probabilities. It doesn't matter in whichever order he tastes, right?



It doesn't and that is why you need to multiply by 4!/2!.

The 9 cups are all distinct - say C1 to C3 have Tea1, C4 to C6 have Tea2 and C7 to C9 have Tea3.

Now if you were select 3 cups, you would do it in 9C3 ways. Then you would arrange them in 3! ways if order mattered.

Note that in our original question when we write (3/9 * 3/8 * 3/7 * 1), we are not just selecting. We are picking a cup with Tea1 (3/9), a cup with Tea2 (3/8) and a cup with Tea3 (3/7) and then any cup (say another Tea2).

So this is the probability of picking Tea1, Tea2, Tea3 and Tea4 in that order only.
But hey, the order does not matter. So Tea2, Tea3, Tea1, Tea2 should be acceptable too and same for all other arrangements of Tea1, Tea2, Tea3 and Tea2. That is why you multiply by 4!/2! (because order DOES NOT matter). So you need to include all arrangements in which one can pick 3 distinct teas. Every such order is acceptable.
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anuj.bit2k10
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Re: At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  11 Aug 2020, 13:06
Bunuel wrote:
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?

# \(\frac{1}{12}\)
# \(\frac{5}{14}\)
# \(\frac{4}{9}\)
# \(\frac{1}{2}\)
# \(\frac{2}{3}\)


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.


Bunuel, \(C^2_3\) - # of ways to chose these 2 cups from the chosen sample - doesn't this consider each cup within a "flavour" different from the others of the same flavour?

If yes, then could you elaborate the reasoning behind considering them different and not same-
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altairahmad
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At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  Updated on: 24 Sep 2020, 04:28
Total number of outcomes:
Number of ways to pick the first cup = 9
Number of ways to pick the second cup = 8
Number of ways to pick the third cup = 7
Number of ways to pick the forth cup = 6
Total number of ways (while adjusting for repetitions) = (9x8x7x6)/4! = 126

Outcomes in which two cups are of same type and two cups are of different types :
Number of ways to pick the first cup (of first type) = 9
Number of ways to pick the second cup (of second type) = 6
Number of ways to pick the third cup (of third type)= 3
Number of ways to pick the forth cup (It can be of any one of the other remaining types) = 6
Total number of ways (while adjusting for repetitions) = (9x6x3x6)/(4!/2!) = 81

Probability of all three types = 81/126 = 9/14

Probability as required = 1 - 9/14 = 5/14

Originally posted by altairahmad on 17 Aug 2020, 05:48.
Last edited by altairahmad on 24 Sep 2020, 04:28, edited 1 time in total.
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Jane121393
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At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  30 Aug 2020, 08:52
Hi VeritasKarishma could you explain for me why my solution is wrong?
My approach: 2 ways for the contestant can't taste all 3 samples: He tastes 3 cups of 1 samples and 1 cup of another sample OR he tastes 2 cups of each 2 samples.
- 1st way: choose 1 sample out of 3 to provide 3 cups: 3C1. Then, 1 sample out of the remaining 2 to provide 1 cup => 2C1 * 3C1
=> 3C1 * 2C1 * 3C1 = 18
- 2nd way: choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2. Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> 3C1 * 3C2 * 2C1 * 3C2 = 54
=> total probability = (18+54)/ 9C4 = 4/7

I can't understand what's wrong with my approach. Please help. Thanks so much.
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At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  30 Aug 2020, 09:14
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Jane121393 wrote:
Hi VeritasKarishma could you explain for me why my solution is wrong?
My approach: 2 ways for the contestant can't taste all 3 samples: He tastes 3 cups of 1 samples and 1 cup of another sample OR he tastes 2 cups of each 2 samples.
- 1st way: choose 1 sample out of 3 to provide 3 cups: 3C1. Then, 1 sample out of the remaining 2 to provide 1 cup => 2C1 * 3C1
=> 3C1 * 2C1 * 3C1 = 18
- 2nd way: choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2. Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> 3C1 * 3C2 * 2C1 * 3C2 = 54
=> total probability = (18+54)/ 9C4 = 4/7

I can't understand what's wrong with my approach. Please help. Thanks so much.


Hi Jane121393


For your 2nd way, you are actually duplicating the count. Lets take an example below:->
Assuming there are there three samples namely S1, S2, and S3. The tea blowing to each on them be S1T1, S1T2, S1T3, S2T1, etc.
S1T1 means Tea cup of 1st variety from sample 1. similarly others.
Now considering your 1st step for way 2 - [b]choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2
You choose sample S1 and then T1 and T2 cup, so you chose S1T1 and S1T2.
In your second step - Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> - you choose sample S2 from remaining S2 and S3 and then 2 cups from it , assuming you choose S2T1 and S2T2.
So finally you have S1T1, S1T2, S2T1 and S2T2.

Now if you choose sample S2 in your 1st step and Cups 1 and 2 -b][b]choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2,
You will get S2T1 and S2T2
And if you choose sample S1 in your 2nd step and then cups 1 and 2 in your 2nd step - [b]Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2 - you will have S1T1 and S1T2.
So finally again you have S2T1 and S2T1 and S1T1 and S1T1. The only difference is that the order in which you selected is different. But here in selection, the order is irrelevant.
So effectively you are taking the same case twice.
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Re: At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  02 Sep 2020, 04:45
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Jane121393 wrote:
Hi VeritasKarishma could you explain for me why my solution is wrong?
My approach: 2 ways for the contestant can't taste all 3 samples: He tastes 3 cups of 1 samples and 1 cup of another sample OR he tastes 2 cups of each 2 samples.
- 1st way: choose 1 sample out of 3 to provide 3 cups: 3C1. Then, 1 sample out of the remaining 2 to provide 1 cup => 2C1 * 3C1
=> 3C1 * 2C1 * 3C1 = 18

Correct.

Jane121393 wrote:
- 2nd way: choose 1 sample out of 3 to provide 2 cups: 3C1 * 3C2. Then, 1 sample out of the remaining 2 to provide 2 cup => 2C1 * 3C2
=> 3C1 * 3C2 * 2C1 * 3C2 = 54
=> total probability = (18+54)/ 9C4 = 4/7

I can't understand what's wrong with my approach. Please help. Thanks so much.


When you choose 3C1 and then 2C1, you are doing it from the same bunch so you are double counting. You need to directly select 2 of the 3. In your case, say you select T1 first and then T3. You take this as one case. Next time, you select T3 first and then T1. You take this an another case. But they both are the same because there is no arrangement.

So this is what you do:
You choose 2 samples out of the 3 samples which will be tasted in 3C2 ways.
For each of the 2 samples, you select 2 of the 3 cups in 3C2 * 3C2 ways.
You get 3C2 * 3C2 * 3C2 = 27 ways


Total = (18 + 27) / 9C4 = 5/14
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Re: At a blind taste competition a contestant is offered 3 cups of each of [#permalink] New post  07 Sep 2020, 12:03
Can someone please explain to me what is wrong with this approach ?

I am calculating the probability of the opposite event - we need to make sure we select atleast 1 cup from each sample -

Options for selecting a cup from the first sample - 3C1
Options for selecting a cup from the second sample - 3C1
Options for selecting a cup from the third sample - 3C1
Options for selecting the fourth cup - 6C1 (as there are 6 options remaining in total)

So req probability = 1- (3C1 * 3C1 * 3C1 * 6C1)/9C4........which is -ve so obviously wrong
Request assistance please
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