Difference between revisions of "2017 AMC 8 Problems/Problem 10"
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− | P (no 5)= <math>\frac{4}{5}</math> * <math>\frac{3 | + | 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. | 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. | ||
Revision as of 23:54, 7 November 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)= *
*
=
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)
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.