2017 AMC 8 Problems/Problem 10
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
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.