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.
I did it by same method. But didn't understand one thing. I took it 1-3C1*3C1*3C1*6C1/9C4. Firstly we can select 1 out of the given three smaples of a particular taste. Then one out of second taste of three. then one out of the remaning three. Now we are left with ^ sampples and we can select any of them and it may fulfil the criterion. But answer is coming wrong. Where am I going wrong. Please help. @veritasprepkarishma
01 Sep 2021, 04:36
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