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

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==Solution 2 (regular probability)==
 
==Solution 2 (regular probability)==
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P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3}{4}</math> * <math>\frac{2}{3}</math> = <math>\frac{2}{5}</math> this is the fraction of total cases with no fives.
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P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3}{4}</math> * <math>\frac{2}{3}</math> = <math>\frac{2}{5}</math>. This is the fraction of total cases with no fives.
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p (no 4 and no 5)= <math>\frac{3}{5}</math> * <math>\frac{2}{4}</math> * <math>\frac{1}{3}</math> = <math>\frac{6}{60}</math> = <math>\frac{1}{10}</math> this is the intersection of no fours and no fives. Subtract fraction of no fours and no fives from no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10}</math> (C)
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p (no 4 and no 5)= <math>\frac{3}{5}</math> * <math>\frac{2}{4}</math> * <math>\frac{1}{3}</math> = <math>\frac{6}{60}</math> = <math>\frac{1}{10}</math>. This is the intersection of no fours and no fives. Subtract the fraction of no fours and no fives from that of no fives. <math>\frac{2}{5} - \frac{1}{10} = \frac{3}{10} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math>.
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==Solution 3 (Complementary Probability)==
 
==Solution 3 (Complementary Probability)==
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Let's have three "boxes."
 
Let's have three "boxes."
 
One of the boxes must be 4, so <math>\frac{\binom{3}{1} \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \boxed{\textbf{(C)}\frac{3}{10}}</math>.
 
One of the boxes must be 4, so <math>\frac{\binom{3}{1} \cdot 3 \cdot 2}{5 \cdot 4 \cdot 3} = \boxed{\textbf{(C)}\frac{3}{10}}</math>.
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 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/P-K9AEAuhNY
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 +
~Education, the Study of Everything
  
 
==Video Solutions==
 
==Video Solutions==
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~savannahsolver
 
~savannahsolver
  
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*https://www.youtube.com/watch?v=F935tcVcXvY
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*https://www.youtube.com/watch?v=F935tcVcXvY   ~David
  
 
==See Also:==
 
==See Also:==

Latest revision as of 19:16, 15 April 2023

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 3 (Complementary Probability)

Using complementary counting:

$P_\textbf{4 is largest} = 1-P_\textbf{5 is largest} - P_\textbf{3 is largest} = 1- \frac{\dbinom{4}{2}}{\dbinom{5}{3}} - \frac{\dbinom{2}{2}}{\dbinom{5}{3}} = 1- \frac{6}{10} - \frac{1}{10} = \boxed{{{\textbf{(C)}\frac{3}{10}}}}$

-mathfan2020

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

~Education, the Study of Everything

Video Solutions

https://youtu.be/FN9qkU62a9U

~savannahsolver

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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