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

1 2 3 4

CrazyBlender123

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15 Jul 2026, 08:58

The numbers are 2,3,5,7,11,13,17,19,23,29,31
The total number of ways 4 numbers can be selected is 11C4 = 330.
For range = 12,
We can choose (5,17), (7,19) , (11,23), (17, 29), (19,31).
When we chose the 1st 3 pairs mentioned above.. there are 3 numbers in between which can be selected in 3C2 ways.
For the remaining two pairs, there are 2 numbers in between which can be selected in 2C2 ways.
So, the total Favourable outcomes are = (3 * 3C2) + (2 * 2C2) = 11
Therefore,
P=11/330 = 1/30
Option B is correct

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Theeldertwin

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15 Jul 2026, 09:03

First 11 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
Sets of numbers which have range of 12: (5, 17) , (7, 19), (11, 23), (17, 29), (19, 31) and the reverse of each of them as well. That gives us 10 favourable outcomes.

Total outcomes= 11P2 or (11C2 × 2!) = ((11×10)/2)×2 = 110

So Probablity = 10/110 = 1/11

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Rishm

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15 Jul 2026, 09:11

I found out, only 5 pairs have a difference of 12, now fav/unfavorable would be 11/4!7!=330 unfav. Now numerator would be how many selections inside the range of 12, which is 3+3+3+1+1 which would be 11/330 and 1/30 which is B is our answer.

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heyaa

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15 Jul 2026, 09:18

11 prime numbers: 2,3,5,7,11,13,17,19,23,29,31.
pairs which work:
5,17 - 7,11,13 = 3c2 =3
7,19 = 3
11,23 =3
17,29 = 19,23 = 2c2 = 1
19,31= 1
total favorable outcomes = 3+3+3+1+1 = 11
total outcomes = 11c4 = 330
answer = 11/330 = 1/30

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RDM42

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15 Jul 2026, 09:21

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?

2,3,5,7,11,13,17,19,23,29,31
We can choose 4 out of 11 without replacement= 11C4 ways = 330 ways
Ranger= Max-Min=12
(5,7,11,13,17) Range=12
Ways of choosing other 2 no.=3C2=3
(7,11,13,17,19) Range= 12
choosing other 2 no.=3C2=3
(11,13,17,19,23) R=12
choosing other 2 no.=3C2=3
(17,19,23,29) R=12
choosing other 2 no.=2C2=1
(19,23,29,31) R=1w
Other 2 no. can be choosed=2C2=1
Valid outcomes= 3+3+3+1+1= 11
P=Valid outcomes/Total outcomes= 11/330=1/30

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yashsharma1

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15 Jul 2026, 09:23

First 11 distinct prime numbers: 2,3,5,7,11,13,17,19,23,29,31
Total ways of picking 4 numbers out of 11 (without replacement) = 11C4 = 330
The question is saying that for favorable outcomes, the range of numbers should be 12. So we need to find how many such cases are possible. For range, we are only concerned with the highest and lowest chosen numbers and the other two can be selected from the remaining possibilities.

The possible combinations are as follows:

1. Highest number = 31, lowest has to be 19. The other two numbers have to be between 19 and 31 and in the list above, there are only 2 such numbers (23 & 29). Hence only 1 combination in this case.
2. Highest = 29, lowest = 17. Other two can only be 19 & 23. Hence 1 possible combination.
3. Highest = 23, lowest = 11. There are 3 numbers (13,17,19) out of which we need to choose the remaining 2. So, 3C2 = 3 possible combinations.
4. Highest = 19, lowest = 7. Again out of 3 (11,13,17) 2 need to be selected. 3C2 = 3 possible combinations.
5. Highest = 17, lowest = 5. We need to choose 2 numbers out of 3 (7,11,13). 3C2 = 3 combinations.

We cannot make a case of 13 and below being the highest number because it will need 1 and below as the lowest number which are not a part of the list.
So, favorable / total = (1+1+3+3+3)/330 = 11/330 = 1/30

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2448

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15 Jul 2026, 09:42

total = 11c4 = 330
possible outcomes = 3c2 + 3c2 + 3c2 + 2c2 + 2c2 = 11
(5,17)(7,19)(11,23)(17,29)(19,31)
3 3 3 2 2
prob = 11/330 = 1/30

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okHedwig

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15 Jul 2026, 09:54

IMO B

the Prime nos : 2,3,5,7,11,13,17,19,23,29,31
Here the nos with range 12 = (5,17), (7,19), (11,23), (17,29), (19,31)
So these nos must be selected in each selection
Now, ways of selecting (5,17) in 4 slips = I have to select 5 & 17, and any prime no between 5 & 17 = 7,11,13
So, ways of selecting 5,17 = 3C1 =3
Silimarly ways of selecting (7,19) =3
Ways of selecting (11,23) =3
Ways of selecting (17, 29 = 1 Since 19, 23 should also be selected so there is only 1 way
Similarly ways of selecting 19,31 =1
Totaal ways = 3+3+3+1+1 =11
Ways of selecting 4 out of 11 slips = 11C4 =11x30
Hence req Prob = 11/11*30 = 1/30
Hope this helps

Bunuel

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AviNFC

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15 Jul 2026, 09:54

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

For range 12:
5,x,y,17..of 7,11,13 any two can be chosen in 3C2=3
7,x,y,19...3C2 = 3ways for x, y
11,x,y,23..3C2 = 3 ways
17,x,y,29..2C2=1 way
19,x,y,31..2C2=1 way

favorable ways= 3+3+3+1+1=11 ways
total ways = 11C4=330

Reqd Prob = 11/330 = 1/30

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Vivek1707

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15 Jul 2026, 10:10

The prime nos are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31

The pairs that will give me a range of 12 will be

31 - 19 ---> 1 way of picking 4 values i.e 19 , 23 , 29 , 31
29 - 17 ---> similarly 1 way
23 - 11 ---> here there ll be 3 ways i.e 1 way for 23 and 11 and choose 2 out of 3 i.e 13 , 17 and 19 --> 1 x 1 x 3 = 3 ways
19 - 7 ---> similarly 3 ways
17 - 5 ---> simlarly 3 ways

Total = 1 + 1 + 3 + 3 + 3 = 11 ways
Sample 11C4 = 330

probabilty 11/330 i.e 1/30 I ll choose B

Bunuel

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sonit1987

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15 Jul 2026, 10:15

Given 11 slips of paper: each has label of first 11 prime numbers.

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

4 slips are drawn randomly one by one without replacement,

probability of all drawn 4 slips range being 12 = ???

Total ways of drawing 4 slips from 11 without replacement= 11C4= (11*10*9*8)/24=330

for range to be 12, the drawn slip must have pairs of (5,17), (7,19), (11,23),(17,29) and (19,31)- 5 pairs
For Range to be 12, we can choose 1 pair out of 5 and remaining two slips label should have value between the selected pair

if (5,17) pair is chosen, then 7, 11, 13 can be chosen for other 2 number label in 3c2 ways= 3 ways
if (7,19) chosen, then 11, 13, 17 can be chosen for other two labels in 3c2 ways= 3 ways
if(11,23) chosen, then 13, 17, 19 to be chosen for other two labels in 3c2 ways= 3 ways
if (17, 29) chosen, then 19, 23 to be chosen for other two label in 1 ways.
if (19,31) chosen, then 23, 29 to be chosen for other two labels in 1 ways

Total= 3+3+3+1+1= 11 ways

Required probability= 11/330= 1/30

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GulfTube

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15 Jul 2026, 10:16

list out first 11 prime numbers: 2,3,5,7,11,13,17,19,23,29,31

to get range of 12 in the picked 4 slips, max-min = 12
so we look for max,min pairs that give range of 12
17,5
19,7
23,11
29,17
31,19

for (17, 5) the other two numbers have to be b/w them otherwise the range will not be 12. so they can be 7,11,3
so prob is 1/11 * 1/10 * 3/9 * 2/8 * 4! = 1/55

similarly for (19,7) the other two numbers can be from 11,13,17
so prob = 1/55
this is also the case for (23,11)

for (29,17) the remaining two numbers can be 19,23 so prob = 1/11*1/10*2/9*1/8*4! = 1/165. same is case for (31,19)

total probability = 3(1/55)+2(1/165) = 1/30
B

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twinkle2311

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15 Jul 2026, 10:30

The first 11 primes are :
2,3,5,7,11,13,17,19,23,29,31

We need 2 primes whose difference is 12 : (5,17) (7,19) (11,23) (17,29) (19,31)

For each pair, choose the remaining 2 primes between them
-> (5,17) : choose 2 from (7,11,13) = 3 ways
-> (7,19) : choose 2 from (11,13,17) = 3 ways
-> (11,23) : choose 2 from (13,17,19) = 3 ways
-> (17,29) : only (19,23) = 1 way
-> (19,31) : only (23,29) = 1 way

Total favorable= 11
Total ways = 11C4 = 330
Probability = 11/330 = 1/30

Ans : B

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MANASH94

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15 Jul 2026, 11:07

1st 11 prime numbers are:
2,3,5,7,11,13,17,19,23,29,31
Pairs for which ranges will be 12 are: (5,17), (7,19), (11,23), (17,29), (19,31)
For 5-17 other prime numbers can be 7, 11, 13... Total ways I can choose 2 numbers out of 3 is 3C2 = 3
For 7-19 => 11,13,17... Total no of ways we can choose 2 numbers is 3C2 = 3
For 11-23 => 13, 17, 19. Total ways to choose 2 numbers is again 3C2 = 3
For 17-29 => 19, 23 similarly here we can choose 1 way
For 19-31 => 23, 29 here we can choose 1 way

Total ways: 3 + 3 + 3 +1 + 1
= 11 ways
Total possibilities is 11C4 =330.
So Probability range of 4 numbers drawn is 12 is 11/330 = 1/30.
Ans is B.

Bunuel

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jefferyillman

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15 Jul 2026, 11:21

First 11 prime numbers are

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

4 numbers choosen without replacement
what are changes of range being 12 = largest - smallest

Prime numbers with a difference of 12 are

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

Therefore there are 5 possible pairs that give 12

for 5,17 you also need either two of 7,11,or 13, 3 choose 2 = 3

for 7,19 you also need either two of 11, 13, 17, 3 choose 2 =3

for 11,23 you also need either two of 13, 17, 19, 3 choose 2 =3

for 17,29 you also need either two 19,23, 2 choose 2 = 1

for 19, 31 you also need either two 23 and 29, 2 choose 2 =1

There favourable sets are 3 + 3 +3 + 1 + 1 =11

All possible sets are 11 select 4 = 330

Probability is therefore 11/330 = 1/30

Answer B 1/30

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Nsanghavi

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15 Jul 2026, 12:10

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

The total outcomes when we select 4 chits frpom 11 chits = 11C4 = 11!/4!*7! = 11*10*9*8/4*3*2*1 = 11*10*3 = 330

The required outcome = Pairs with Range of 12

Total pairs with range of 12 are as below:
17, 5, any 2 other numbers between 17 and 5 - i.e 3 ways
19, 7 and any 2 numbers between 19 and 7 - i.e. 3 ways
11, 23 and any 2 numbers between 11 and 23 - i.e. 3 ways
17, 29 and any 2 numbers between 17 and 29 - i.e. 1 way
19, 31 and any 2 numbers between 19 and 21 - i.e. 1 way
= 3*3 +2 = 11

Hence the probability = 11/330 = 1/30 - Option B

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Shlok02

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15 Jul 2026, 12:14

Bunuel

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first 11 Prime numbers = 2,3,5,7,11,13,17,19,23,29,31
For a set of 4 numbers selected with the range 12 we will have the follwoing sets
Set 1 - 5, x, y, 17. We can select 2 out of 7,11 and 13 = 3C2 = 3 ways
Set 2- 7, x,y, 19. We can select 2 out of 11,13 and 17 = 3C2 = 3 ways
Set 3- 11,x,y,23. We can select 2 out of 13,17,19 = 3c2 = 3 ways
Set 4 - 17,19,23,29. Only 1 way
Set 5 - 19,23,29, 31. Only 1 way

Total sets = 3 + 3 + 3 + 1 +1 = 11
Prob = 11/ 11c4 = 1/30 (B)

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chasing725

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15 Jul 2026, 12:33

Bunuel

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1. Between 5 & 17, we need to select two numbers = 3C2 = 3
2. Between 11 & 23, we need to select two number = 3C2 = 3
3. Between 7 & 19, we need to select two number = 3C2 = 3
4. Between 17 & 29, we need to select two number = 2C2 = 1
5. Between 19 & 31, we need to select two number = 2C2 = 1

Total = 11

11/11C4 = 11/11*3 * 10 = 1/30

Option B

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SmileAndSolve

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15 Jul 2026, 13:30

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FlushLoop

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15 Jul 2026, 14:06

Bunuel

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

Pairs with difference 12:

5 and 17 = 3 choices for the other 2 numbers
7 and 19 = 3 choices
11 and 23 = 3 choices
17 and 29 = 1 choice
19 and 31 = 1 choice

Total favorable = 11

Total ways: C(11,4) = 330

Probability: 11/330 = 1/30

Option B seems correct.

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