Difference between revisions of "1986 AHSME Problems/Problem 22"
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The total number of ways to choose 6 numbers is <math>{10\choose 6} = 210</math> | The total number of ways to choose 6 numbers is <math>{10\choose 6} = 210</math> | ||
| − | Assume 3 is the lowest number. There are 5 numbers left to choose, each of which has to be greater than 3. This is equivalent to choosing 5 numbers from the 7 numbers larger than 3, <math>{7\choose | + | Assume 3 is the lowest number. There are 5 numbers left to choose, each of which has to be greater than 3. This is equivalent to choosing 5 numbers from the 7 numbers larger than 3, <math>{7\choose 5} = 21</math> |
Thus, <math>\frac{21}{210} = \frac{1}{10}</math> <math>\fbox{E}</math> | Thus, <math>\frac{21}{210} = \frac{1}{10}</math> <math>\fbox{E}</math> | ||
Revision as of 00:56, 2 September 2015
Problem
Six distinct integers are picked at random from 
. What is the probability that, among those selected, the second smallest is 
?
Solution
The total number of ways to choose 6 numbers is 
Assume 3 is the lowest number. There are 5 numbers left to choose, each of which has to be greater than 3. This is equivalent to choosing 5 numbers from the 7 numbers larger than 3, 
Thus, 
See also
| 1986 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
