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

Revision as of 07:48, 5 March 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 Solutions

https://youtu.be/FN9qkU62a9U

~savannahsolver

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

@@ Line 10: / Line 10: @@
 ==Solution 2 (regular probability)==
-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.
+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.
-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)
+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}</math>= \boxed{{{\textbf{(C)}\frac{3}{10}}}}$.
 ==Solution 3 (Complementary Probability)==

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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