Difference between revisions of "2017 AMC 12B Problems/Problem 22"
(→Solution) |
(→Solution) |
||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | It amounts to filling in a 4x4 | + | It amounts to filling in a 4x4 matrix M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let <RA>=number of non-0 elements in RA. |
− | WOLG, let C1=[1,-1,0,0]. Parity demands that <RA> and <RB> = 1 or 3. | + | WOLG, let C1=[1,-1,0,0]. Parity demands that <RA> and<RB>= 1 or 3. |
CASE 1. <RA>=3, and <RB>=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here. | CASE 1. <RA>=3, and <RB>=3. 3 choices for -1in RA, then everything is determined. So 3 subcases here. |
Revision as of 12:06, 17 February 2017
Problem 22
Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?
Solution
It amounts to filling in a 4x4 matrix M. Columns C1 - C4 are the random draws each round; rows RA -RD are the coin changes of each player. Also, let <RA>=number of non-0 elements in RA.
WOLG, let C1=[1,-1,0,0]. Parity demands that <RA> and
CASE 1. <RA>=3, and
Case 2. <RA>=1, and
Case 3. <RA>=1, and
In sum, 3+24+18=45 cases. Probability= 45÷(12^3)=5/192.
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.