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14 Jul 2026, 08:00
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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. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 391/400
A. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 391/400
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M46-44
14 Jul 2026, 09:00
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GMAT Club 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. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 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(Red) = 5/10 = 1/2
P(Blue) = 3/10
For one specific order, such as Red, Red, Blue, Blue, the probability is:
1/2 * 1/2 * 3/10 * 3/10 = 9/400
The 2 Red choices and 2 Blue choices can be arranged in:
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 * 9/400 = 54/400 = 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 - 27/200 = 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. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 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(Red) = 5/10 = 1/2
P(Blue) = 3/10
For one specific order, such as Red, Red, Blue, Blue, the probability is:
1/2 * 1/2 * 3/10 * 3/10 = 9/400
The 2 Red choices and 2 Blue choices can be arranged in:
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 * 9/400 = 54/400 = 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 - 27/200 = 173/200
Answer: D.
General Discussion
14 Jul 2026, 08:11
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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. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 391/400
total contestants are 10
3 teams Blue , red & green
red has 5 , green has 2 and blue
total rounds of game are 4
P of getting red blue & green members
.5 , .3 , .2
and not getting .5 , .7 , .8
not exactly 2 are of red and 2 of blue
total arrangements 4c2 ; 6
exactly two red & two blue
6*.5*.4 * .5*.4 = 0.135
P of not getting will be 1-0.135= 0.865
173/200 is correct option D
A. 27/200
B. 1/7
C. 6/7
D. 173/200
E. 391/400
total contestants are 10
3 teams Blue , red & green
red has 5 , green has 2 and blue
total rounds of game are 4
P of getting red blue & green members
.5 , .3 , .2
and not getting .5 , .7 , .8
not exactly 2 are of red and 2 of blue
total arrangements 4c2 ; 6
exactly two red & two blue
6*.5*.4 * .5*.4 = 0.135
P of not getting will be 1-0.135= 0.865
173/200 is correct option D
14 Jul 2026, 08:23
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P ( Red ) =5/10
p ( blue )=3/10
p(green)=2/10=1/5
so , p ( 2 blue , 2 red) = 5.5.3.3/10.10.10.10= 9/100
2 red and 2 blue can be arranged in 4p2 ways = 4! /2! =6 way
probability of getting 2 red and 2 blue = 6 . 9/100 = 54/400 =27/200
probability of not getting 2 red and 2 blue = (1 - 27/200 )
= 173/200
so answer is 173/200
p ( blue )=3/10
p(green)=2/10=1/5
so , p ( 2 blue , 2 red) = 5.5.3.3/10.10.10.10= 9/100
2 red and 2 blue can be arranged in 4p2 ways = 4! /2! =6 way
probability of getting 2 red and 2 blue = 6 . 9/100 = 54/400 =27/200
probability of not getting 2 red and 2 blue = (1 - 27/200 )
= 173/200
so answer is 173/200
