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15 Jul 2026, 08:00
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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
A. 1/66
B. 1/30
C. 1/22
D. 1/15
E. 1/11
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M46-45
15 Jul 2026, 09:00
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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.
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.
General Discussion
15 Jul 2026, 08:17
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Here's my solution for this question
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15 Jul 2026, 08:17
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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
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
