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GMAT Club World Cup 2026 (Day 7): At a quiz show, 10 contestants are

1 2 3 4

GurpreetMahli

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Answer: B = 1/7

Given: 5 reds, 3 blues, 2 greens, 4 dependent picks

Need: Probability, in 4 picks, of exactly 2 reds and 2 blues

Combination method because when we select for order it takes to account of possible order of picks so we don't have to use permutations.

Step 1. Total possible outcomes

10 contestants and 4 picks = 10 chose 4 = 10 * 3 * 7 ~ keep it factored out for easier division later on

Step 2. Favorable outcomes

Exactly 2 reds and 2 blues

(Red) 5 chose 2 = 5 * 2

(Blue) 3 chose 2 = 3

And = *

So 5 * 2 * 3

Step 3. Favorable Outcomes divided by Total Outcomes

(5 * 2 * 3 )/ (10 * 3 * 7 ) = 1/7

Bunuel

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

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Total possibilities is equal 10 C 4. 10!/6!4! which is equal to 210.

Now we have to determine every possibility except for one when there exactly 2 people from team red and two from team blue. The easiest way to do this is to subtract the odds of that specific outcome from 1.

For Red, we have 5 C 2 which equals ten. For blue, we have 3 c 2 which equals 3. This gives us total of 30 possible options where there are exactly two red and two blue contestants.

210/210 - 30/210 = 180/210 =6/7. OPTION C IMO

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Answer:

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As contestants can be chosen again in a latter round, this is a probability with replacement problem.
We have to calculate 1 - p(choosing 2 from Red and 2 from Blue in the four rounds), hence let us calculate this probability first.

Red has 5, and Blue has 3. The two chosen from Red could be the same person, and the two chosen from Blue could be the same person due to replacement.
We can calculate this as p(chosen from Red)*p(chosen from Red)*p(chosen from Blue)*p(chosen from Blue)*[(4!)/(2!2!)] = (5C1/10C1)*(5C1*10C1)*(3C1*10C1)*(3C1*10C1)*[(4!)/(2!2!)]=27/200

Thus the desired probability is 1-(27/200)=173/200

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Total contestants are 10.
so P(1 red) = 5/10=1/2
P(1 green) = 2/10= 1/5
P(1 blue) = 3/10

now we want to find exactly 2 R and 2 Blue and then we can negate it.
P(RRBB) = (1/2 * 1/2* 3/10*3/10)= 9/400

Now arrangement of RRBB is 4c2 => 6
So total combination is 6*(9/400) = 27/200

So our answer would be 1 - P(2R and 2B)
=1-(27/200) => 173/200

Answer is D

Bunuel

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Given Team Members (Contestants):
Red (R) = 5
Blue (B) = 3
Green (G) = 2

There are 4 rounds and in each round, 1 Contestant is chosen at random from all the 10.
One constestant chosen in 1 round can be again chose for another round.

The question is asking for the probability that the 4 constestants are not exactly 2 Rs and 2 Bs.
It is always better to apply complement rule here. So first, find the probability of the 4 contestants chosen to be exactly 2Rs and 2Bs.

We know there are 5 reds. Any of them can be chosen for one round. Similarly for another round too (because we want two reds out of 4 contestants)
Likewise, we know there are 3 blues. Any of them can be chosen for 2 rounds.
So the numerator would be 5 * 5 * 3 * 3
But, we they can be in any sequence of 4. RRBB, RBBR, etc so multiply the above with 4! / 2! (R and B is repeating)

Overall outcomes/events - any of the 10 can be chosen for any of the round and are eligible to repeat. So 10^4

So the final probability of having exactly 2 reds and exactly 2 blues = [(5^2 * 3^2) / 10^4] * 4!/2! = 27 / 200

So the probability of not having the arrangement of exactly 2 reds and 2 blues = 1 - 27/200 = 173 / 200. Answer is D.

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For each round, getting a red contestant is P(R) = 5/10; and a blue contestant, P(B) = 3/10
Getting 2 Reds out of 4 rounds will be 4c2 = 6
Exactly 2 red and 2 blue = 6 x (5/10 x 5/10) x (3/10 x 3/10) = 6 x 1/2 x 1/2 x 9/100 = 27/200
Now, since the question is asking not exactly 2 red and 2 blue, the probability will be 1 - (27/200) = 173/200

Answer: D

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Given,
P (Red) = 5/10 = 1/2
P (Blue) = 3/10
P (Green) = 2/10 = 1/5

Probability ( exactly 2 Red & 2 Blue) = 4C2 * (1/2)^2 * (3/10)^2
= 6*(1/4)*(9/100)
= 54/400
= 27/200

We need,
Probability (Not exactly 2 Red & 2 Blue) = 1 - Probability (Exactly 2 Red & 2 Blue)
= 1 - (27/200)
= 173/200

Answer : D (173/200)

Bunuel

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Probability for choosing contestants from Red Team is 5/10 and Team Blue is 3/10.
This question is like choosing ball with replacement from a box.
Possible ways of choosing 4 contestants with repeatation RRBB with arrangements in 4!/2!*2!
= 6 x 5/10 x 5/10 x 3/10 x 3/10
= 27/200

Probability of not having exactly 2 from Red and 2 from blue team will be = 1-27/200 = 173/200 D

Bunuel

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The answer is D. 173/200
Because each round is independent because contestants stay eligible. In any single round, the chance of picking someone from team red is five out of ten, which is one half. The chance of picking someone from team blue is three out of ten.
First find the probability of the outcome we want to avoid: exactly 2 red picks and exactly 2 blue picks across the four rounds. The 2 red rounds can be arranged among the four rounds in six ways, since four choose two equals six. Each specific arrangement has probability 1/2x1/2×3/10×3/10, which is 9/400. Multiplied by six ways to get 54/400, which simplify to 27/200.
The question ask for the probability that this does not happen, so subtract from one: 1-27/200 equals 173/200. That is choice D.

Bunuel

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5R , 3 b, 2 g , P of not having exaclty 2 R and 2 b
out of 10 , 1 selected and remain eligible for next round
total no of outcomes = 10 raise to 4

Probablity of selecting red in pone round = 5/10 = 1/2

Prpbaltiy of sleecting blue in one round = 3/10

Probalty of exactly 2r and 2 b = 1/2 * 1/2 * 3/10 * 3/10 * (4!/2!*2! ways)
= 1/4*9/100* (4*3/2 ways)
= 9/400*6 = 27/200

Problity of not selecting exaclty 2 r and 2 b = 1-27/200 = 173/200

option d

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The probability of picking someone on the red team is 5/10, the probability of green is 2/10, and the probability of blue is 3/10 all based on # of members on team/ total contestants.

The probability that the 4 rounds do not consist of 2 red and 2 blue is equal to 1 - P(2 red and 2 blue). The number of different orders the 2 red and 2 blue is equal to 4!/(2!2!) = 6. Then I multiplied the probability of R squared (because it's picked twice) which is (1/2)^2 = 1/4 then multiply by the probability of B squared (3/10)^2 = 9/100. After multiplying I got 27/200. Since we are trying to find the probability it is not 2 red and 2 blue, I subtracted from 1 to get 173/200. D is the answer.

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P(Red)=5/10=0.5
P(Blue)=3/10=0.3
No. of ways to arrange 2R and 2B over 2 rounds is 4C2=6.
P(RRBB)=0.5*0.5*0.3*0.3=0.0225=9/400
P(Not Exactly 2R and 2B) = 1 - (4C2*P(RRBB))
= 1 - 27/200 = 173/200
The solution is D.

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there are 3 teams
red= 5 people'
blue= 3 people
green = 2 people.

there are 4 rounds. and repetition is allowed here.
P(no 2 from red and 2 from blue).

we can find the opposite of it thats 2 from red and 2 from blue and then subtract them from 1.

so we have RRBB.
number of ways to have this = 4! / 2!2! = 6 ways.

6* (5/10)(5/10)(3/10)(3/10)

27/200

1- 27/200

173/200

choice D

Bunuel

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10 contestants total
red-5, blue-3, green-2
Each of the 4 rounds selects 1 contestant from all 10, replacing them.
I need the probability that the 4 choices are not made up of exactly 2 red and 2 blue.
Probability of exactly 2 reds and 2 blues.
Pr=5/10=1/2, Pb=3/10

to get 2 reds and 2 blues in 4 rounds
choose which 2 of the 4 are red(4/2)=6
Probability of any specific arrangement:
(1/2)^2(3/10)^2=1/4*9/100=9/400
hence P(x 2 red, 2 blue)=6*9/400=54/100=27/200
Prob(not exactly 2 r and 2 b)=1-27/200=173/200
ANS: D. 173/200

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contestants remaining eligible each round is an independent choice from all 10 contestants.
P(Red) = 5/10 P(Blue) = 3/10

as the question is a negative one, i.e. that this does not happen so best to follow 1 - P(does happen)

first calculate props of getting exactly 2 red and 2 blue

choose which 2 of the 4 rounds contain red 4 chose 2 = 6

For R,R,B,B = 0.5^2 x 0.3^2

Probs of 2 Red and 2 Blue = 6x0.5^2 x 0.3^2 = 27/200

then dont forget to minus, 1-27/200 = 173/200 = D

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I used the technique of subtracting from total probability of 1.
i found prob of 2 red & 2 blue contestants = 5/10 * 5/10 * 3/10 * 3/10 * 4!/2! = 27/200

prob of not exactly 2 red & 2 blue = 1 - 27/200 = 173/200
D

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1. Total ways of picking 4 contestants = Since replacement is allowed, in each of the round, 10 options are available = 10^4
2. Picking 2 contestants from the red team for 2 out of 4 rounds = 5*5
3. Picking 2 contestants from the blue team for 2 out of 4 rounds = 3*3

2 & 3 above give us the options available for picking the desired contestants. But these contestants can be picked in the following 6 ways

Round 1(R1), Round 2(R2), Round 3(R3), Round 4(R4)
A. RRBB
B. RBRB
C. RBBR
D. BBRR
E. BRBR
F. BRRB

This can also be calculated as 4!/2!2! = 6

Final answer = 1- probability of exactly 2 contestants being from red and blue team each = 1 - (5*5*3*3*6 / 10^4) = 1- (1350/10000) = 1- (27/200) = 173/200

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P(Red) = 5/10 = 1/2
P(Blue) = 3/10
P(Green) = 2/10
We need the probability that the 4 selections are NOT made up of exactly 2 contestants from Team Red and exactly 2 contestants from Team Blue

Finding the probability of getting exactly 2 Red and 2 Blue:

Choose which 2 of the 4 rounds are Red: 4C2 = 6

Probability of any one arrangement:
-----> (1/2)2 × (3/10)2
-----> (1/4) × (9/100)
-----> 9/400

So, P(exactly 2 Red and 2 Blue) = 6 × 9/400
-----> 54/400
-----> 27/200

P(Not exactly 2 Red and 2 Blue)
= 1 − 27/200
= 173/200

Answer: D

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two important thinks to do in this first solve for individual probalility

easy way to do that is to find opposite of what you need . so for that we will get TR=0.5 and TB=0.3

now also we need probability of exactly 2 blue and 2 red right
we use permutation here as order doesent matter and that will be 4!/2!*2!
answer will be 6 for this
now we need probability of each individual out come which is 6*0.5^2*0.3^2=0.135
now the last step is probability of what we need as 0.135 is probabilty of what we dont want
then we do that by 1-0.135=0.865 which we translate in fraction we get 173/200