2024 AMC 12B Problems/Problem 25
Contents
[hide]Problem 25
Pablo will decorate each of identical white balls with either a striped or a dotted pattern, using either red or blue paint. He will decide on the color and pattern for each ball by flipping a fair coin for each of the decisions he must make. After the paint dries, he will place the balls in an urn. Frida will randomly select one ball from the urn and note its color and pattern. The events "the ball Frida selects is red" and "the ball Frida selects is striped" may or may not be independent, depending on the outcome of Pablo's coin flips. The probability that these two events are independent can be written as where and are relatively prime positive integers. What is (Recall that two events and are independent if )
Solution 1
Let be the number of balls that are both striped and red, be the number of balls that are striped but blue, and be the number of balls that are red but dotted. Then there must be balls that are dotted and blue.
Let be the event, "the ball Frida selects is red", and be the event, "the ball Frida selects is striped". and are independent if and only if , i.e.,
Before we continue, I'd like to clarify that the sample space , under which we are computing probability that and are independent, consists of total outcomes, each equally likely. Notice that once given , there are corresponding outcomes.
Now, it remains to find all nonnegative integer solutions to the above equation such that , add up all the corresponding outcomes, and then divide by . We can find the solutions relatively easily by doing casework on . Long story short, one finds the solutions to be
There are little tricks we can use to make the process of adding up the corresponding outcomes faster. First, we notice that the first, second, and last row of solutions contributes outcomes. Then we can take advantage of the symmetry between to add up the rest of the outcomes. In the end we get
Hence, the probability is . Therefore, the answer is
~tsun26
Solution 2 Symmetry + Observations = 5 min solve
Let be the number of balls with stripes, the number of red balls, and the number of balls that are red and have stripes. Now, , or .
Make the trivial observation that whenever or equals , or both equal , there is one unique satisfying the equality. Notice for and going from to , we have a total probability of due to binomial theorem. By symmetry, we must multiply by to account for switching and , and another for the case where . Finally, we subtract the four overlapping cases and , where each pair has format . Thus, the total probability is .
Now, there are more pairs that satisfy the condition, namely and .
Again, make the trivial observation that each of these have the same probability due to invariance and similar distribution.
For a single pair, take as an example. We know , and we fix . The probability is therefore .
Adding all of them up, we get . Thus, the answer is .
Remarks - You could also solve this problem using floor/ceiling states and the Gaussian-Eulerian Method by isolating the floor function.
~Apollo Luo, edited by Mathemagician108
See also
2024 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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.