2000 AIME I Problems/Problem 5
Problem
Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is 
One marble is taken out of each box randomly. The probability that both marbles are black is 
and the probability that both marbles are white is 
where 
and 
are relatively prime positive integers. What is 
?
Solution
If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate 
. The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.
Let 
represent the number of marbles in each box, and without loss of generality let 
. Then, 
, and since the 
may be reduced to form 
on the denominator of 
, 
. It follows that 
, so there are 2 pairs of 
and 
.
- Case 1: Then the product of the number of black marbles in each box is
, so the only combination that works is 
black in first box, and 
black in second. Then, 
so 
.
, so the only combination that works is 
black in first box, and 
black in second. Then, 
so 
.- Case 2: The only combination that works is 9 black in both. Thus,
. .
. Thus, 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
