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16 Sep 2014, 00:13
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In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?
A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
16 Sep 2014, 00:13
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Official Solution:
In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?
A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
To ensure that the contestant does not taste all three types of tea, he must taste only two types of tea. Notice that it's impossible for him to taste just one type of tea, since he will be tasting four cups and in any case, will have to try more than one type of tea.
The number of ways to taste only two types of tea is:
\(\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}\):
\(C^2_3\) is the number of ways to choose which 2 types of tea out of 3 will be tasted;
\(C^4_6\) is the number of ways to choose 4 cups out of 6 cups from these 2 types of tea (2 types = 6 cups);
\(C^4_9\) is the total number of ways to choose 4 cups out of 9.
Another way:
Calculate the probability of opposite event and subtract this value from 1.
The opposite event will be for the contestant to taste ALL 3 samples, so the contestant must choose 2 cups from one type of tea, and 1 cup from each of the other two types (2-1-1).
\(C^1_3\) is the number of ways to choose the type of tea which will provide with 2 cups;
\(C^2_3\) is the number of ways to choose these 2 cups from the chosen type;
\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the second type;
\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the third type;
\(C^4_9\) is the total number 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
In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?
A. \(\frac{1}{12}\)
B. \(\frac{5}{14}\)
C. \(\frac{4}{9}\)
D. \(\frac{1}{2}\)
E. \(\frac{2}{3}\)
To ensure that the contestant does not taste all three types of tea, he must taste only two types of tea. Notice that it's impossible for him to taste just one type of tea, since he will be tasting four cups and in any case, will have to try more than one type of tea.
The number of ways to taste only two types of tea is:
\(\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}\):
\(C^2_3\) is the number of ways to choose which 2 types of tea out of 3 will be tasted;
\(C^4_6\) is the number of ways to choose 4 cups out of 6 cups from these 2 types of tea (2 types = 6 cups);
\(C^4_9\) is the total number of ways to choose 4 cups out of 9.
Another way:
Calculate the probability of opposite event and subtract this value from 1.
The opposite event will be for the contestant to taste ALL 3 samples, so the contestant must choose 2 cups from one type of tea, and 1 cup from each of the other two types (2-1-1).
\(C^1_3\) is the number of ways to choose the type of tea which will provide with 2 cups;
\(C^2_3\) is the number of ways to choose these 2 cups from the chosen type;
\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the second type;
\(C^1_3\) is the number of ways to choose 1 cup out of 3 from the third type;
\(C^4_9\) is the total number 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
22 Jul 2018, 10:04
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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
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
