Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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<math>\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}</math> | <math>\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}</math> | ||
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==Solution== | ==Solution== | ||
− | 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{{\ | + | 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{{\dbinom{3}{2}}}{{\dbinom{5}{3}}}} = \boxed{{\textbf{(C) }} {\frac{3}{10}}}</math> |
==Solution 2 (regular probability)== | ==Solution 2 (regular probability)== | ||
− | P (no 5)= 4/ | + | 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 | + | 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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+ | ==Solution 3 (Complementary Probability)== | ||
+ | Using complementary counting, <math>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{{\frac{3}{10}}{\textbf{(C)}}}</math> | ||
+ | -mathfan2020 | ||
+ | ==Solution 4== | ||
+ | Let's have three 'boxes'. | ||
+ | One of the boxes must be 4, so 3C1 x 3 x 2/5 x 4 x 3 = 3/10 | ||
− | Video | + | ==Video Solutions== |
− | https://youtu.be/M9kj4ztWbwo | + | *https://youtu.be/OOdK-nOzaII?t=1237 |
+ | *https://youtu.be/M9kj4ztWbwo | ||
==See Also:== | ==See Also:== |
Revision as of 02:48, 29 September 2021
Contents
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?
Solution
There are possible groups of cards that can be selected. If is the largest card selected, then the other two cards must be either , , or , for a total groups of cards. Then the probability is just
Solution 2 (regular probability)
P (no 5)= * * = this is the fraction of total cases with no fives. p (no 4 and no 5)= * * = = this is the intersection of no fours and no fives. Subtract fraction of no fours and no fives from no fives. (C)
Solution 3 (Complementary Probability)
Using complementary counting, -mathfan2020
Solution 4
Let's have three 'boxes'. One of the boxes must be 4, so 3C1 x 3 x 2/5 x 4 x 3 = 3/10
Video Solutions
See Also:
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 |
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