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17 Jul 2026, 15:52
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
32% (02:01) correct
68%
(02:28)
wrong
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A fair six-sided die is rolled three times. What is the probability that the range of the numbers that turn up is 3?
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{9}\)
D. \(\frac{1}{4}\)
E. \(\frac{1}{3}\)
A. \(\frac{1}{9}\)
B. \(\frac{1}{6}\)
C. \(\frac{2}{9}\)
D. \(\frac{1}{4}\)
E. \(\frac{1}{3}\)
17 Jul 2026, 21:08
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Oh I like this one, "range" questions with dice always trip people up because everyone wants to jump straight to counting outcomes without setting up a system first.
1. Range = max roll minus min roll. We need range = 3, meaning if the min is m, the max has to be exactly m+3. Since dice only go up to 6, m can only be 1, 2, or 3 (windows: 1-4, 2-5, 3-6).
2. Total outcomes: since we're rolling the same die three separate times, order matters, so it's 6^3 = 216, not some combination count. This is where I've seen people go wrong, they treat it like choosing 3 values from a set instead of 3 independent rolls.
3. For one window, say 1 to 4, I need all three rolls to land somewhere in {1,2,3,4}, but the min actually has to hit 1 and the max actually has to hit 4 (otherwise the range would be smaller than 3). That's an inclusion-exclusion setup: all three in {1,2,3,4} is 4^3=64. Subtract the ones that never hit 1 (all in {2,3,4}, so 3^3=27) and the ones that never hit 4 (all in {1,2,3}, 27 again). Then add back the ones missing both, since we subtracted those twice (all in {2,3}, 2^3=8).
4. So valid triples for one window = 64 - 27 - 27 + 8 = 18.
5. Same math applies to the other two windows (2-5 and 3-6), so total = 18 x 3 = 54.
6. Probability = 54/216 = 1/4.
Answer is D.
The trap here really is the inclusion-exclusion step, it's so tempting to just do 4^3 and stop, or forget you're double subtracting. I got burned by almost the identical setup on a min/max question in practice and just plugged 3^3 without the correction term. Once you see it as "count the whole window, then carve out the middle," it clicks.
1. Range = max roll minus min roll. We need range = 3, meaning if the min is m, the max has to be exactly m+3. Since dice only go up to 6, m can only be 1, 2, or 3 (windows: 1-4, 2-5, 3-6).
2. Total outcomes: since we're rolling the same die three separate times, order matters, so it's 6^3 = 216, not some combination count. This is where I've seen people go wrong, they treat it like choosing 3 values from a set instead of 3 independent rolls.
3. For one window, say 1 to 4, I need all three rolls to land somewhere in {1,2,3,4}, but the min actually has to hit 1 and the max actually has to hit 4 (otherwise the range would be smaller than 3). That's an inclusion-exclusion setup: all three in {1,2,3,4} is 4^3=64. Subtract the ones that never hit 1 (all in {2,3,4}, so 3^3=27) and the ones that never hit 4 (all in {1,2,3}, 27 again). Then add back the ones missing both, since we subtracted those twice (all in {2,3}, 2^3=8).
4. So valid triples for one window = 64 - 27 - 27 + 8 = 18.
5. Same math applies to the other two windows (2-5 and 3-6), so total = 18 x 3 = 54.
6. Probability = 54/216 = 1/4.
Answer is D.
The trap here really is the inclusion-exclusion step, it's so tempting to just do 4^3 and stop, or forget you're double subtracting. I got burned by almost the identical setup on a min/max question in practice and just plugged 3^3 without the correction term. Once you see it as "count the whole window, then carve out the middle," it clicks.
17 Jul 2026, 21:36
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Total # outcomes when the die is thrown 3 times = 6 x 6 x 6
Favorable outcomes:
Logic
- The range of the numbers is 3.
- The numbers obtained, when arranged in ascending order are a, _, a+3.
- What could the middle term be? Anything from a to a+3, both included (a, a+1, a+2, a+3)
There are only a few possibilities for what these numbers can be ->
(1) 1, 1/2/3/4, 4
(2) 2, 2/3/4/5, 5
(3) 3, 3/4/5/6, 6
No other possibilities.
But also remember - we also need to count the possible arrangements of numbers within every possible case above.
For example --- (1, 2, 4) is different from (4, 2, 1). Our total outcomes (6 x 6 x 6) counts these two as separate. So, when calculating favorable outcomes, we need to count all possible arrangements too.
Calculation
a, _, a+3
(1) How many ways to select "a"? 3 (Because "a" can only be 1 or 2 or 3).
(2) Once the lowest number "a" is chosen, there is only 1 possibility for the highest number "a+3".
(3) Based on which middle term is selected, the number of arrangements varies. So, let's break this down.
a) If the middle term is "a", then the numbers obtained are (a, a, a+3). There are 3!/2! = 3 total arrangements.
b) If the middle term is "a+3", then the numbers obtained are (a, a+3, a+3). There are 3!/2! = 3 total arrangements.
c) If the middle term is "a+1", then the numbers obtained are (a, a+1, a+3). There are 3! = 6 total arrangements.
d) If the middle term is "a+2", then the numbers obtained are (a, a+2, a+3). There are 3! = 6 total arrangements.
Either of the above 4 could happen. So, the total possibilities = 3 + 3 + 6 + 6 = 18.
#Favorable Outcomes = (1) x (2) x (3) = 3 x 1 x 18
Probability = Favorable/Total = \(\frac{(3 * 18) }{ (6 * 6 * 6)}\) = 1/4. Choice D.
---
Harsha
Favorable outcomes:
Logic
- The range of the numbers is 3.
- The numbers obtained, when arranged in ascending order are a, _, a+3.
- What could the middle term be? Anything from a to a+3, both included (a, a+1, a+2, a+3)
There are only a few possibilities for what these numbers can be ->
(1) 1, 1/2/3/4, 4
(2) 2, 2/3/4/5, 5
(3) 3, 3/4/5/6, 6
No other possibilities.
But also remember - we also need to count the possible arrangements of numbers within every possible case above.
For example --- (1, 2, 4) is different from (4, 2, 1). Our total outcomes (6 x 6 x 6) counts these two as separate. So, when calculating favorable outcomes, we need to count all possible arrangements too.
Calculation
a, _, a+3
(1) How many ways to select "a"? 3 (Because "a" can only be 1 or 2 or 3).
(2) Once the lowest number "a" is chosen, there is only 1 possibility for the highest number "a+3".
(3) Based on which middle term is selected, the number of arrangements varies. So, let's break this down.
a) If the middle term is "a", then the numbers obtained are (a, a, a+3). There are 3!/2! = 3 total arrangements.
b) If the middle term is "a+3", then the numbers obtained are (a, a+3, a+3). There are 3!/2! = 3 total arrangements.
c) If the middle term is "a+1", then the numbers obtained are (a, a+1, a+3). There are 3! = 6 total arrangements.
d) If the middle term is "a+2", then the numbers obtained are (a, a+2, a+3). There are 3! = 6 total arrangements.
Either of the above 4 could happen. So, the total possibilities = 3 + 3 + 6 + 6 = 18.
#Favorable Outcomes = (1) x (2) x (3) = 3 x 1 x 18
Probability = Favorable/Total = \(\frac{(3 * 18) }{ (6 * 6 * 6)}\) = 1/4. Choice D.
---
Harsha
