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14 Apr 2026, 17:50
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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:
65%
(hard)
Question Stats:
42% (01:43) correct
58%
(02:46)
wrong
based on 24
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If 3 of the 10 smallest positive integers are chosen at random, what is the probability that the median of the chosen integers is 5 or 6?
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{10}\)
D. \(\frac{1}{3}\)
E. \(\frac{2}{5}\)
A. \(\frac{1}{5}\)
B. \(\frac{1}{4}\)
C. \(\frac{3}{10}\)
D. \(\frac{1}{3}\)
E. \(\frac{2}{5}\)
15 Apr 2026, 04:49
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The answer is D (1/3).
The concept being tested here is Probability combined with counting - specifically, you need to figure out what combinations of 3 integers give a median of 5 or 6.
The 10 smallest positive integers are 1 through 10. When you pick 3 of them, the median is just the middle value when arranged in order.
1. Total ways to choose 3 from 10: C(10,3) = 10!/(3! x 7!) = 120
2. Ways to get median = 5: You need exactly one number below 5 and one above 5, with 5 in the middle.
- Numbers below 5: {1, 2, 3, 4} -- 4 choices
- Numbers above 5: {6, 7, 8, 9, 10} -- 5 choices
- Favorable for median = 5: 4 x 5 = 20
3. Ways to get median = 6: Same logic, 6 must be the middle.
- Numbers below 6: {1, 2, 3, 4, 5} -- 5 choices
- Numbers above 6: {7, 8, 9, 10} -- 4 choices
- Favorable for median = 6: 5 x 4 = 20
4. Total favorable: 20 + 20 = 40
5. P(median = 5 or 6) = 40/120 = 1/3
The trap most people fall into is trying to enumerate cases or applying inclusion-exclusion when you don't need to. Median = 5 and median = 6 are mutually exclusive events, so you can just add directly.
Also worth noting: the symmetry here is beautiful and not an accident. The integers 5 and 6 are the two middle numbers of the set 1-10, so by symmetry each has the same count (20). If you spot this during the exam you can save a calculation.
Answer: D
The concept being tested here is Probability combined with counting - specifically, you need to figure out what combinations of 3 integers give a median of 5 or 6.
The 10 smallest positive integers are 1 through 10. When you pick 3 of them, the median is just the middle value when arranged in order.
1. Total ways to choose 3 from 10: C(10,3) = 10!/(3! x 7!) = 120
2. Ways to get median = 5: You need exactly one number below 5 and one above 5, with 5 in the middle.
- Numbers below 5: {1, 2, 3, 4} -- 4 choices
- Numbers above 5: {6, 7, 8, 9, 10} -- 5 choices
- Favorable for median = 5: 4 x 5 = 20
3. Ways to get median = 6: Same logic, 6 must be the middle.
- Numbers below 6: {1, 2, 3, 4, 5} -- 5 choices
- Numbers above 6: {7, 8, 9, 10} -- 4 choices
- Favorable for median = 6: 5 x 4 = 20
4. Total favorable: 20 + 20 = 40
5. P(median = 5 or 6) = 40/120 = 1/3
The trap most people fall into is trying to enumerate cases or applying inclusion-exclusion when you don't need to. Median = 5 and median = 6 are mutually exclusive events, so you can just add directly.
Also worth noting: the symmetry here is beautiful and not an accident. The integers 5 and 6 are the two middle numbers of the set 1-10, so by symmetry each has the same count (20). If you spot this during the exam you can save a calculation.
Answer: D
