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07 Aug 2026, 05:09
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At a quiz show, 10 contestants are divided into 3 teams: Team Red has 5 contestants, Team Blue has 3 contestants, and Team Green has 2 contestants. The show has 4 rounds, and in each round, 1 contestant is chosen at random from all 10 contestants. If a contestant chosen in one round remains eligible to be chosen again in any later round, what is the probability that the 4 choices are not made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue?
A.
\(\frac{27}{200}\)
B.
\(\frac{1}{7}\)
C.
\(\frac{6}{7}\)
D.
\(\frac{173}{200}\)
E.
\(\frac{391}{400}\)
A.
\(\frac{27}{200}\)
B.
\(\frac{1}{7}\)
C.
\(\frac{6}{7}\)
D.
\(\frac{173}{200}\)
E.
\(\frac{391}{400}\)
07 Aug 2026, 05:09
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Official Solution:
At a quiz show, 10 contestants are divided into 3 teams: Team Red has 5 contestants, Team Blue has 3 contestants, and Team Green has 2 contestants. The show has 4 rounds, and in each round, 1 contestant is chosen at random from all 10 contestants. If a contestant chosen in one round remains eligible to be chosen again in any later round, what is the probability that the 4 choices are not made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue?
A.
\(\frac{27}{200}\)
B.
\(\frac{1}{7}\)
C.
\(\frac{6}{7}\)
D.
\(\frac{173}{200}\)
E.
\(\frac{391}{400}\)
First find the probability that the 4 choices are made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue.
Since the selection is with replacement, the probabilities stay the same in every round. So, in any one round:
\(P(\text{Red}) = \frac{5}{10} = \frac{1}{2}\)
\(P(\text{Blue}) = \frac{3}{10}\)
For one specific order, such as Red, Red, Blue, Blue, the probability is:
\(\frac{1}{2} * \frac{1}{2} * \frac{3}{10} * \frac{3}{10} = \frac{9}{400}\)
The 2 Red choices and 2 Blue choices can be arranged in:
\(\frac{4!}{2!2!} = 6\) ways
So the probability that the 4 choices are made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue is:
\(6 * \frac{9}{400} = \frac{54}{400} = \frac{27}{200}\)
Therefore, the probability that the 4 choices are not made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue is:
\(1 - \frac{27}{200} = \frac{173}{200}\)
Answer: D
At a quiz show, 10 contestants are divided into 3 teams: Team Red has 5 contestants, Team Blue has 3 contestants, and Team Green has 2 contestants. The show has 4 rounds, and in each round, 1 contestant is chosen at random from all 10 contestants. If a contestant chosen in one round remains eligible to be chosen again in any later round, what is the probability that the 4 choices are not made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue?
A.
\(\frac{27}{200}\)
B.
\(\frac{1}{7}\)
C.
\(\frac{6}{7}\)
D.
\(\frac{173}{200}\)
E.
\(\frac{391}{400}\)
First find the probability that the 4 choices are made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue.
Since the selection is with replacement, the probabilities stay the same in every round. So, in any one round:
\(P(\text{Red}) = \frac{5}{10} = \frac{1}{2}\)
\(P(\text{Blue}) = \frac{3}{10}\)
For one specific order, such as Red, Red, Blue, Blue, the probability is:
\(\frac{1}{2} * \frac{1}{2} * \frac{3}{10} * \frac{3}{10} = \frac{9}{400}\)
The 2 Red choices and 2 Blue choices can be arranged in:
\(\frac{4!}{2!2!} = 6\) ways
So the probability that the 4 choices are made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue is:
\(6 * \frac{9}{400} = \frac{54}{400} = \frac{27}{200}\)
Therefore, the probability that the 4 choices are not made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue is:
\(1 - \frac{27}{200} = \frac{173}{200}\)
Answer: D
