2017 AMC 8 Problems/Problem 10

Problem 10

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?

$\textbf{(A) }\frac{1}{10}\qquad\textbf{(B) }\frac{1}{5}\qquad\textbf{(C) }\frac{3}{10}\qquad\textbf{(D) }\frac{2}{5}\qquad\textbf{(E) }\frac{1}{2}$

Solution

There are $5 \choose 3$ possible groups of cards that can be selected. If $4$ is largest card selected, then the other two cards must be either $1$ , $2$ , or $3$ , for a total $3 \choose 2$ groups of cards. Then the probability is just ${\frac{{3 \choose 2}}{{5 \choose 3}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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2017 AMC 8 Problems/Problem 10

Problem 10

Solution

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