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GMAT Club World Cup 2026 (Day 8): A box contains 11 slips of paper.

1 2 3 4

Bunuel

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Bunuel

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GMAT Club Official Solution:

A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip. If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?

A. 1/66
B. 1/30
C. 1/22
D. 1/15
E. 1/11

The range is the positive difference between the greatest chosen number and the smallest chosen number.

The first 11 prime numbers are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

For the range to be 12, the smallest and greatest chosen numbers must differ by 12.

The possible endpoint pairs are:

5 and 17
7 and 19
11 and 23
17 and 29
19 and 31

For each pair, the two endpoints are fixed, and there is only 1 way to choose that pair. The remaining 2 numbers must be chosen from the primes strictly between those endpoints.

For 5 and 17: choose 2 from 7, 11, 13, giving C(3,2) = 3 ways.
For 7 and 19: choose 2 from 11, 13, 17, giving C(3,2) = 3 ways.
For 11 and 23: choose 2 from 13, 17, 19, giving C(3,2) = 3 ways.
For 17 and 29: choose 2 from 19, 23, giving C(2,2) = 1 way.
For 19 and 31: choose 2 from 23, 29, giving C(2,2) = 1 way.

So the number of favorable selections is:

3 + 3 + 3 + 1 + 1 = 11

The total number of ways to choose 4 slips from 11 is:

C(11,4) = 330

Therefore, the probability is:

11/330 = 1/30

Answer: B.

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Here's my solution for this question

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A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip. If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?

A. 1/66
B. 1/30
C. 1/22
D. 1/15
E. 1/11

first 11 prime numbers are
2,3,5,7,11,13,17,19,23,29,31

condition is 4 slips are drawn in random from 11 slips

total ways 11c4 = 330 ways
outcome possible that range is 12 possible when pair of prime is

(5,7) ; (7,19) ; (11,23) ; (17,29); ( 19,31)

in between these pairs the prime numbers which are there can be choosen
for ( 5,7) ; ( 7,19) ; ( 11,23)
are 3 numbers
3c2 ways for each ; 3+3+3 ; 9
and for (17,29) ; ( 19,31)
2c1 : 1 + 1 = 2
9+2 = 11

total ways possible 11/330

1/30 option B is correct option

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15 Jul 2026, 08:18

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total ways 11C4=330
need range=12
valid endpoint pairs
5,17: 3c2=3
7,19:(3c2)=3
11,23:(3C2)=3
17,29:2c2=1
19,31: 2C2=1

3+3+3+1+1=11

11/330=1/30

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First 11 primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
Pairs that gives difference of 12= (5,17),(7,19),(11,23),(17,29),(19,31)
Range needs to be 12. So the lowest and highest number must be from pair.
Ways of choosing other 2 number between (5,17)= 3C2=3
.............(7,19)= 3C2 =3
............(11,23)= 3C2 =3
............(17,29)= 2C2 =1
............ (19,31)= 2C2 =1
Total ways pf choosing 4 numbers that have range 4= 3+3+3+1+1=11
Ways of selecting any 4 out of 11 primes = 11C4 =11*10*9*8/4*3*2=330
Required Probability = 11/330 =1/30

B. 1/30

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First 11 prime numbers = 2,3,5,7,11,13,17,19,23,29,31

If the range is 12, then pairs of {min, max} = {5,17}, {7,19}, {11,23}, {17,29}, {19,31}

For each range, possible outcomes are:
{5,17} = 3c2 =3
{7,19} = 3c2 =3
{11,23} =3c2=3
{17,29} = 2c2 =1
{19,31} = 2c2 =1

Total favourable outcomes = 3+3+3+1+1 =11

total outcomes = 11C4 = 330
Required probabilty = 11/330 = 1/30 (Ans)

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A box contains 11 slips of paper. The slips are labeled with the first 11 prime numbers, with one different prime number on each slip.

If 4 slips are drawn at random from the box without replacement, what is the probability that the range of the 4 numbers drawn is 12?

First 11 prime numbers = {2,3,5,7,11,13,17,19,23,29,31}

12 = 17 - 5 = 19 - 7 = 23 - 11 = 29 - 17 = 31 - 19

Total ways to choose 4 numbers out of 11 = 11C4 = 330

Favorable ways:
Case 1: 12 = 17-5
We have to chose 2 numbers out of {7,11,13} = 3C2 = 3
Case 2: 12 = 19-7
We have to chose 4 numbers out of {11,13,17} = 3C2 = 3
Case 3: 12 = 23-11
We have to chose 4 numbers out of {13,17,19} = 3C2 = 3
Case 4: 12 = 29-17
We have to chose 4 numbers out of {19,23} = 2C2 = 1
Case 5: 12 = 31-19
We have to chose 4 numbers out of {23,29} = 2C2 = 1

Favorable ways = 3+3+3+1+1 = 11

If 4 slips are drawn at random without replacement, the probability that the range of 4 numbers drawn is 12 = 11/330 = 1/30

IMO B

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Answer: B= 1/30

the first 11 primes are:

2,3,5,7,11,13,17,19,23,29,31

The range is difference between the smallest and the largest numbers which should differ by 12.

valid pairs:
(5,17) , (7,19) , (11,23) , (17,29) , (19,31)

the primes between each of them:
(7,11,13)= 3
(11,13,17)= 3
(13,17,19)=3
(19,23)=1
(23,29)=1

total fav sets: 3+3+3+1+1= 11

total ways to choose 4 slips: 11c4= 1/30

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1st 11 prime no:
2,3,5,7,11,13,17,19,23,29,31

range of 12
17,5
19,7
11,23
17,29
19,31

for each, other 2 numbers must lie strictly between endpoints
5,17 -> 7,11,13
7,19 -> 11,13,17
11,23 -> 13,17,19
17,29 -> 19,23
19,31 -> 23,29

Favorable selections: 3+3+3+1+1 =11
Total ways to choose 4 slips: 11C4 = 330
Hence 11/330
1/30
Option :B

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Total possibilities 11C4 = 330
First 11 prime numbers include = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
For the difference to be 12, the numbers have to be greater than 13 at least as minimum prime number is 2
17-5, 19-7, 23-11, 29-17, 31-19

The other 2 numbers will strictly need to be between these numbers.
Till the 23,11 pair we have 3 primes between highest and lowest. Thus, ways of selection = 3 x 3C2 = 3 x 3 = 9
After 23,11 pair we have 2 primes between highest and lowest. Thus, ways of selection = 2 x 2C2 = 2 x 1 = 2

Prob = 11/330 = 1/30 (B)

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Total ways of selecting = 11c4 = 330

there are 5 pairs for which we get a range of 12

5,17 / 7,19 / 11,23 / 17,29 / 19,31

there are 3 nos between 5,17 / 7,19, / 11, 23 and 2 bw 17,29 / 19, 31

so 3c2 * 3 + 2c2 * 2 = 11 = 11/330 = 1/30.

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the First 11 prime numbers and keeping them as min value for the range of 12, the max value should be

P - 2,3,5,7,11,13,17,19,23,27,29,31
M - 14,15,17,19,23,25,29,31 and so on

so possible Combos with - _Min_ ___ ___ _Max_
5 _ _ 17 - The center two digits has to be within these limits and can be filled only in ascending order in 3C2 (7,11,13) ways = 3

5 7 11 17
5 7 13 17
5 11 13 17
3 ways

7 _ _ 19 - similarly 3
11 _ _ 23 - 3
17 _ _ 29 - 1 because only one case is possible 17 19 23 29
19 _ _ 31 - 1 again (19 23 29 31)

3+3+3+1+1 = 11

Total ways = 11C4

so probability = 11/ 11C4 = 11*4*3*2*1/ 11*10*9*8 = 1/30

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i am going with option b
as there are 330 ways of choosing 4 slips
only 11 groups have a range of exactly 12
thus, 11/330 = 1/30
a incorrect - this shows 5 favourable groups but there are 11.
c incorrect - this would show 15 favourable groups.
d incorrect - this would show 22 fav groups
e incorrect - this would require 30 fav groups

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first 11 prime numbers = 2 ,3 ,5,7,11,13,17,19,23,29,31

number of ways to choose 4 from 11 = 330

total pair with a range of 12 = 5 &17
7&19
11 ,23
17,29
19,31

there are 3 prime numbers between ( 5,17) , (7,19) ,(11,23) .so here , n(n-1)/2 = 3.2/3 = 3 ways to pick 2 number from each one .
similiarly , 1 is the only way to choose 2 prime number from ( 17,29) and (19,31) .

total favourable situation = 3.3 + 2 = 11

Probability of desired outcome =11/330 =1/30

ans is 1/30

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Answer: B) 30

First 11 primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
After adding 12: 14,15,17,19,23, 25, 29, 31, 35, 41, 43
Possible pairs: x, x, x, x, x

There are 5 possible pairs; however, not every pair has the same number of ways to satisfy the range requirement.
Pair (5,17): has 3 prime numbers between 5 and 7 to choose 2 from 3C2 = 3
Pair (7,19): 3C2 = 3
Pair (11,23): 3C2 = 3
Pair (17,29): only has 2 prime numbers between 17 and 29 so there is only one possible way
Pair (19,31): same as Pair (17,29)

Probability = total favourable outcomes / total possible outcomes = (3+3+3+1+1) / 11C4 = 11 / 330 = 1/30

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The first step I did was writing all first 11 prime numbers:
2,3,5,7,11,13,17,19,23,29,31

Then I checked what pair should I subtract to get 12

Pair 1: 17-5, three values are between 17 and 5[which are 7,11,13] and we need only two. So probability would be (3C2)/(11C4)

Pair 2: 19-7, similarly for this too, the probability would be same as pair 1's

Pair 3: 23-11, this would also have same probability as pair 1's

Pair 4: 29-17, we have only two values between them i.e. 19 and 23. So the probability would be (2C2)/(11C4)

Pair 5: 31-19, this would have probability similar to Pair 4's

So now we have got all probabilities let us add them up, (3+3+3+1+1)/(11C4)
After solving I directly marked 1/22.

If you have any confusions then let me know I will reply at my earliest

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let S be the set of the first 11 primes. So, S = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}
Now, each pair with a range of 12 = {5,17}, {7,19}, {11,23}, {17,29}, {19,31}

Now, for the {5, 17} pair, since all the numbers will be within 5 & 17 so be within the range, let the subset be S1 = {5, 7, 11, 13, 17}
We know that in the 5 and 17 subset, 5 & 17 must be in the group to have 12 as a range, so the remaining 2 numbers out of 3 can be chosen in 3C2 ways = 3 ways.

Similarly, for {7, 19} & {11, 23} subsets, we have 3 ways of choosing.

Now for {17, 29} and {19, 31} pairs, they have 4 elements each. So there is only 1 way to chose elements in them.

Hence, the total number of favourable cases = 3 + 3 + 3 + 1 + 1 = 11
Total number of cases = 11C4 = 11 x 30

Hence the correct probablility = 11 / (11 x 30) = 1/30. Answer is B.

Bunuel

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15 Jul 2026, 09:26

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Let's start by listing the first 11 prime numbers in our box:
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31}

First, we calculate the total number of ways to blindly draw 4 slips from these 11.
Total possible draws = $C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$.

We want the range of the 4 drawn numbers to be exactly 12. This means the difference between the largest and smallest number in our specific draw must be 12. Let's find every pair of primes from our list that gives a difference of 12, and see how many ways we can fill in the remaining 2 numbers from the primes trapped between them.

Pair (5, 17): To complete our group of 4, we need 2 more primes strictly between 5 and 17. The available primes are {7, 11, 13}.
Ways to choose 2 = $C(3, 2) =$ 3 possibilities.
Pair (7, 19): Available primes strictly between them are {11, 13, 17}.
Ways to choose 2 = $C(3, 2) =$ 3 possibilities.
Pair (11, 23): Available primes strictly between them are {13, 17, 19}.
Ways to choose 2 = $C(3, 2) =$ 3 possibilities.
Pair (17, 29): Available primes strictly between them are {19, 23}.
Ways to choose 2 = $C(2, 2) =$ 1 possibility.
Pair (19, 31): Available primes strictly between them are {23, 29}.
Ways to choose 2 = $C(2, 2) =$ 1 possibility.

Adding those up, the total number of favorable outcomes is 3 + 3 + 3 + 1 + 1 = 11.

Finally, the probability is simply the favorable outcomes divided by the total possible draws:
$\frac{11}{330} =$ $\frac{1}{30}$

Correct Answer: B

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15 Jul 2026, 09:32

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From first 11 prime nummbers, the pairs that give difference of 12 are:
5,17 7,19 11,23 17,29 19,31 (took me good 1 minute to figure this lol)
Now, for each pair, we have 2 numbers already, the extremes, and we need to select 2 more from prime numbers in between them to complete 4 options.
So,
from 1st pair - 3C2 =3
2nd pair= 3C2 = 3
3rd= same= 3
4th= 2C2 = 1
5th= 2C2 = 1
TOTAL = 3+3+3+1+1 =11
Total outcomes altogether = 11C4 = 330
Now,
Probability = 11/330 = 1/30