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Question Stats:
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A set J consists of 7 consecutive integers. If a subset of J containing at least two integers is selected at random, approximately what is the probability that the subset and J have the same median?
A. 16%
B. 23%
C. 32%
D. 41%
E. 50%
A. 16%
B. 23%
C. 32%
D. 41%
E. 50%
24 Mar 2026, 09:38
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Nice problem. The trap here is getting the denominator wrong - people often forget to exclude single-element subsets, or they miscalculate the even-sized subset cases.
Let J = {n, n+1, n+2, n+3, n+4, n+5, n+6}. Median of J = n+3.
Total subsets with at least 2 elements = 2^7 - 1 (empty) - 7 (singletons) = 120.
Now count subsets whose median also equals n+3.
1. Size 2: median = average of both elements = n+3, so the two elements sum to 2n+6. Pairs: {n, n+6}, {n+1, n+5}, {n+2, n+4}. That's 3.
2. Size 3: n+3 must be the middle element. Pick 1 from {n, n+1, n+2} and 1 from {n+4, n+5, n+6}, include n+3. That's 3 x 3 = 9.
3. Size 4: need (x2 + x3)/2 = n+3, so x2+x3 = 2n+6. Valid pairs for middle two: {n+1, n+5} gives 1 choice each side = 1 subset; {n+2, n+4} gives 2 choices on left, 2 on right = 4 subsets. Total: 5.
4. Size 5: n+3 must be the 3rd element. Choose 2 from the 3 below n+3 and 2 from the 3 above. C(3,2) x C(3,2) = 9.
5. Size 6: exclude exactly one element. The only way the median (avg of 3rd and 4th) equals n+3 is if you exclude n+3 itself, leaving {n, n+1, n+2, n+4, n+5, n+6} where the middle two are n+2 and n+4, averaging to n+3. That's 1 subset.
6. Size 7: that is J itself, median = n+3. Count 1.
Total favorable: 3 + 9 + 5 + 9 + 1 + 1 = 28.
Probability = 28/120 = 7/30 ≈ 23.3%.
Closest answer is B (22%). The small gap is just rounding - 23.3% sits between 22% and 33%, but it's clearly closer to 22.
The concept tested here is median-by-size in combinations. The even-sized cases (2, 4, 6) where the median is an average trip people up the most - worth knowing cold for GMAT Focus.
Let J = {n, n+1, n+2, n+3, n+4, n+5, n+6}. Median of J = n+3.
Total subsets with at least 2 elements = 2^7 - 1 (empty) - 7 (singletons) = 120.
Now count subsets whose median also equals n+3.
1. Size 2: median = average of both elements = n+3, so the two elements sum to 2n+6. Pairs: {n, n+6}, {n+1, n+5}, {n+2, n+4}. That's 3.
2. Size 3: n+3 must be the middle element. Pick 1 from {n, n+1, n+2} and 1 from {n+4, n+5, n+6}, include n+3. That's 3 x 3 = 9.
3. Size 4: need (x2 + x3)/2 = n+3, so x2+x3 = 2n+6. Valid pairs for middle two: {n+1, n+5} gives 1 choice each side = 1 subset; {n+2, n+4} gives 2 choices on left, 2 on right = 4 subsets. Total: 5.
4. Size 5: n+3 must be the 3rd element. Choose 2 from the 3 below n+3 and 2 from the 3 above. C(3,2) x C(3,2) = 9.
5. Size 6: exclude exactly one element. The only way the median (avg of 3rd and 4th) equals n+3 is if you exclude n+3 itself, leaving {n, n+1, n+2, n+4, n+5, n+6} where the middle two are n+2 and n+4, averaging to n+3. That's 1 subset.
6. Size 7: that is J itself, median = n+3. Count 1.
Total favorable: 3 + 9 + 5 + 9 + 1 + 1 = 28.
Probability = 28/120 = 7/30 ≈ 23.3%.
Closest answer is B (22%). The small gap is just rounding - 23.3% sits between 22% and 33%, but it's clearly closer to 22.
The concept tested here is median-by-size in combinations. The even-sized cases (2, 4, 6) where the median is an average trip people up the most - worth knowing cold for GMAT Focus.
