Difference between revisions of "1986 AHSME Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | <math> | + | 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 18:37, 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 .
Thus, , which is .
See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AHSME Problems and Solutions |
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