Difference between revisions of "1986 AHSME Problems/Problem 22"
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==Solution== | ==Solution== | ||
| − | <math>fbox{ | + | The total number of ways to choose 6 numbers is <math>{10\choose 6} = 210</math>. |
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| + | Assume <math>3</math> is the second-lowest number. There are <math>5</math> numbers left to choose, <math>4</math> of which must be greater than <math>3</math>, and <math>1</math> of which must be less than <math>3</math>. This is equivalent to choosing <math>4</math> numbers from the <math>7</math> numbers larger than <math>3</math>, and <math>1</math> number from the <math>2</math> numbers less than <math>3</math>. <cmath>{7\choose 4} {2\choose 1}= 35\times2.</cmath> | ||
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| + | Thus, <math>\frac{35\times2}{210} = \frac{1}{3}</math>, which is <math>\fbox{C}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 17:37, 12 May 2021
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 is the second-lowest number. There are 
numbers left to choose, 
of which must be greater than , and 
of which must be less than . This is equivalent to choosing numbers from the 
numbers larger than , and number from the 
numbers less than . ![\[{7\choose 4} {2\choose 1}= 35\times2.\]](http://latex.artofproblemsolving.com/8/1/7/8179362f683e40871fc3360361b3e04cd1ea33d2.png)
Thus, 
, which is 
.
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 | ||
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