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

(Solution)
(Solution)
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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 6} = 21</math>   
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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 $\{1,2,3,\ldots,10\}$. 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 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, ${7\choose 5} = 21$

Thus, $\frac{21}{210} = \frac{1}{10}$ $\fbox{E}$

See also

1986 AHSME (ProblemsAnswer KeyResources)
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
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