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Difference between revisions of "2017 AMC 8 Problems/Problem 10"

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There are <math>\binom{5}{3}</math> possible groups of cards that can be selected. If <math>4</math> is the largest card selected, then the other two cards must be either <math>1</math>, <math>2</math>, or <math>3</math>, for a total <math>\binom{3}{2}</math> groups of cards. Then the probability is just <math>{\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math>
 
There are <math>\binom{5}{3}</math> possible groups of cards that can be selected. If <math>4</math> is the largest card selected, then the other two cards must be either <math>1</math>, <math>2</math>, or <math>3</math>, for a total <math>\binom{3}{2}</math> groups of cards. Then the probability is just <math>{\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math>
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==Video Solution (regular probability)==
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https://youtu.be/M9kj4ztWbwo
  
 
==See Also:==
 
==See Also:==

Revision as of 23:37, 4 October 2020

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 ${\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$

Video Solution (regular probability)

https://youtu.be/M9kj4ztWbwo

See Also:

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

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