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15 Jul 2026, 02:09
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Probability for not choosing exactly 2 from Red and 2 from Blue = 1 - Probability for choosing exactly 2 from R and 2 from B
So, to compute probability for exactly 2 from R and 2 from B:
Total outcomes = Total ways of choosing 4 contestants for each of the four rounds = 10C1 * 10C1 * 10C1 * 10C1
(Because we can choose 1 contestant out of all 10 in each round regardless of whether they have been chosen before or not)
Ways to choose exactly 2 from R and 2 from B=
4C2 * 5 * 5 * 3 * 3
(We first choose any two rounds out of 4 in which we will put Red contestants.
Then, for each round chosen, we can choose any one contestant out of the 5, so it will be 5*5 and then for each of the remaining 2 rounds, it will be any 1 contestant out of 3 Blue)
Note that we won't do 5C2 and 3C2 here (I thought through this while solving) because for each round we can choose one out of all the contestants available, so the same person can go for both rounds)
Probability of choosing 2 from Red and 2 from Blue = 6*25*9/10^4 = 27/200
Probability for not choosing this = 1 - 27/200= 173/200
So, to compute probability for exactly 2 from R and 2 from B:
Total outcomes = Total ways of choosing 4 contestants for each of the four rounds = 10C1 * 10C1 * 10C1 * 10C1
(Because we can choose 1 contestant out of all 10 in each round regardless of whether they have been chosen before or not)
Ways to choose exactly 2 from R and 2 from B=
4C2 * 5 * 5 * 3 * 3
(We first choose any two rounds out of 4 in which we will put Red contestants.
Then, for each round chosen, we can choose any one contestant out of the 5, so it will be 5*5 and then for each of the remaining 2 rounds, it will be any 1 contestant out of 3 Blue)
Note that we won't do 5C2 and 3C2 here (I thought through this while solving) because for each round we can choose one out of all the contestants available, so the same person can go for both rounds)
Probability of choosing 2 from Red and 2 from Blue = 6*25*9/10^4 = 27/200
Probability for not choosing this = 1 - 27/200= 173/200
15 Jul 2026, 02:19
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The correct answer. 173/200. D.
Probability of choosing from Red team: 5/10=1/2
Probability of choosing from Blue team: 3/10.
We need to choose 2 from 4 rounds for red contestants.
4C2= 6 ways.
Calculate the probability of 1 specific sequence.
P(R)xP(R)xP(B)xP(b)= (1/2)^2* (3/10)^2= (1/4)*9/100
Multiplying the number of arrangements:
P (exactly 2 red and 2 blue)= 6*9/400= 54/400= 27/200.---(1)
For probability of Not exactly 2 red and 2 Blue we substract (1) from 1-27/200= 173/200.
Probability of choosing from Red team: 5/10=1/2
Probability of choosing from Blue team: 3/10.
We need to choose 2 from 4 rounds for red contestants.
4C2= 6 ways.
Calculate the probability of 1 specific sequence.
P(R)xP(R)xP(B)xP(b)= (1/2)^2* (3/10)^2= (1/4)*9/100
Multiplying the number of arrangements:
P (exactly 2 red and 2 blue)= 6*9/400= 54/400= 27/200.---(1)
For probability of Not exactly 2 red and 2 Blue we substract (1) from 1-27/200= 173/200.
15 Jul 2026, 02:23
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Probability of choosing:
Team Red = 5/10 = 1/2
Team Blue = 3/10
Team Green = 2/10 = 1/5
We want the probability of getting exactly 2 Red and 2 Blue
Choose which 2 of the 4 rounds are Red : 4C2 = 6 Ways
Probability of any 1 arrangement : ((1/2)^2) * ((3/10)^2) = 9/400
So, P(2Red & 2Blue) = 6 * (9/400) = 27/200
Required Prob. = 1-(27/200) = 173/200
Ans : D
Team Red = 5/10 = 1/2
Team Blue = 3/10
Team Green = 2/10 = 1/5
We want the probability of getting exactly 2 Red and 2 Blue
Choose which 2 of the 4 rounds are Red : 4C2 = 6 Ways
Probability of any 1 arrangement : ((1/2)^2) * ((3/10)^2) = 9/400
So, P(2Red & 2Blue) = 6 * (9/400) = 27/200
Required Prob. = 1-(27/200) = 173/200
Ans : D
