Difference between revisions of "2017 AMC 8 Problems/Problem 10"
Tutorjames (talk | contribs) (→Solution) |
Tutorjames (talk | contribs) (→Video Solution (regular probability)) |
||
| Line 12: | Line 12: | ||
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> | ||
| − | == | + | ==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 | https://youtu.be/M9kj4ztWbwo | ||
Revision as of 22:44, 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?
Video Solution
https://youtu.be/OOdK-nOzaII?t=1237
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)= 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 (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
