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?

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

Video Solution

https://youtu.be/OOdK-nOzaII?t=1237

Solution

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)= 4/5*3/4*2/3=2/5 this is the fraction of total cases with no fives. p (no 4 and no 5)= 3/5*2/4*1/3= 6/60=1/10 this is the intersection of no fours and no fives. Subtract fraction of no fours and no fives from no fives.

$\frac{2}{5} - \frac{1}{10} = \frac{3}{10}$ (C)

Video here: https://youtu.be/M9kj4ztWbwo

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

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

Page

Toolbox

Search

2017 AMC 8 Problems/Problem 10

Contents

Problem 10

Video Solution

Solution

Solution 2 (regular probability)

See Also: