2000 AIME I Problems/Problem 5
Contents
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 1
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
.
- Case 2: The only combination that works is 9 black in both. Thus,
.
.
Thus, .
Solution 2
Let and
represent the white and black marbles in boxes 1 and 2.
Since there are marbles in the box:
From the fact that there is a chance of drawing one black marble from each box:
Thinking of the numerator and denominator separately, if was not a reduced fraction when calculating out the probability, then
. Since
, this forces the variables to be
and
in some permutation. Without loss of generality, let
and
.
The denominator becomes:
Since there have been black marbles used, there must be
white marbles. Substituting that in:
Since the factors of that are greater than
are
and
, the quantity
must equal one of those. However, since
, testing
and
for
does not give a correct product. Thus,
must be a reduced form of the actual fraction.
First assume that the fraction was reduced from , yielding the equations
and
.
Factoring
and saying WLOG that
gives
or
. Trying the first pair and setting the denominator equal to 100 gives:
Since , the pairs
and
can be tried, since each box must contain at least one white marble. Plugging in
gives the true equation
, so the number of marbles are
Thus, the chance of drawing 2 white marbles is in lowest terms, and the answer to the problem is
For completeness, the fraction may be tested.
is the highest necessary denominator that needs to be tested, since the maximum the denominator
can be when the sum of all integer variables is
is when the variables are
and
, in some permutation, which gives
. If
, this forces
, since all variables must be integers under
. The denominator becomes
, and since there are now
white marbles total, the denominator becomes
. Testing
gives a solution, and thus
. The complete solution for this case is
. Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is
, which still simplifies to
.
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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