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Difference between revisions of "1986 AHSME Problems/Problem 22"

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==Solution==
 
==Solution==
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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 $\{1,2,3,\ldots,10\}$. What is the probability that, among those selected, the second smallest is $3$?

$\textbf{(A)}\ \frac{1}{60}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{3}\qquad \textbf{(D)}\ \frac{1}{2}\qquad \textbf{(E)}\ \text{none of these}$


Solution

The total number of ways to choose 6 numbers is ${10\choose 6} = 210$.

Assume $3$ is the second-lowest number. There are $5$ numbers left to choose, $4$ of which must be greater than $3$, and $1$ of which must be less than $3$. This is equivalent to choosing $4$ numbers from the $7$ numbers larger than $3$, and $1$ number from the $2$ numbers less than $3$. \[{7\choose 4} {2\choose 1}= 35\times2.\]

Thus, $\frac{35\times2}{210} = \frac{1}{3}$, which is $\fbox{C}$.

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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