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

(Solution)
(Video Solution (regular probability))
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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>
  
==Video Solution (regular probability)==
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==Solution 2 (regular probability)==
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P (no 5)= 4/5*3/4*2/3=2/5 this would be the fraction of total cases with no fives.
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p (no 4 or 5)= 3/5*2/4*1/3= 6/60=1/10
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2/5-1/10=3/10
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Video here:
 
https://youtu.be/M9kj4ztWbwo
 
https://youtu.be/M9kj4ztWbwo
  

Revision as of 23:44, 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 ${\frac{{\binom{3}{2}}}{{\binom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}$

Solution 2 (regular probability)

P (no 5)= 4/5*3/4*2/3=2/5 this would be the fraction of total cases with no fives. p (no 4 or 5)= 3/5*2/4*1/3= 6/60=1/10 2/5-1/10=3/10


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

See Also:

2017 AMC 8 (ProblemsAnswer KeyResources)
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

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