1986 AHSME Problems/Problem 22

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 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. ${7\choose 4} {2\choose 1}= 35*2$

Thus, $\frac{35*2}{210} = \frac{1}{3}$ $\fbox{C}$

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1986 AHSME Problems/Problem 22

Problem

Solution

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