Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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==Solution 4== | ==Solution 4== | ||
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>. |
==Video Solutions== | ==Video Solutions== | ||
Revision as of 07:26, 8 January 2023
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 1 (combinations)
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 
.
Video Solutions
~savannahsolver
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. 
