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 1 (combinations)

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

Solution 2 (regular probability)

P (no 5)= $\frac{4}{5}$ * $\frac{3}{4}$ * $\frac{2}{3}$ = $\frac{2}{5}$ . This is the fraction of total cases with no fives. p (no 4 and no 5)= $\frac{3}{5}$ * $\frac{2}{4}$ * $\frac{1}{3}$ = $\frac{6}{60}$ = $\frac{1}{10}$ . This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. $\frac{2}{5} - \frac{1}{10} = \frac{3}{10} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$ .

Solution 4

Let's have three "boxes." One of the boxes must be 4, so $\frac{\binom{3}{1} \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \boxed{\textbf{(C)}\frac{3}{10}}$ .

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/P-K9AEAuhNY

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Video Solutions

https://youtu.be/FN9qkU62a9U

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https://www.youtube.com/watch?v=F935tcVcXvY ~David

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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