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+ | ==Solution 2 Symmetry + Observations = 5 min solve== | ||
+ | Let <math>S</math> be the number of balls with stripes, <math>R</math> the number of red balls, and <math>N</math> the number of balls that are red and have stripes. | ||
+ | Now, <math>P(A\ \text{and}\ B) = P(A) \cdot P(B) \Rightarrow \frac{N}{6}=\frac{S \cdot R}{36}</math>, or <math>6N = SR</math>. | ||
+ | |||
+ | Make the trivial observation that whenever <math>S</math> or <math>R</math> equals <math>0</math>, or both equal <math>6</math>, there is one unique <math>N</math> satisfying the equality. | ||
+ | Notice for <math>S = 0</math> and <math>R</math> going from <math>0</math> to <math>6</math>, we have a total probability of <math>\frac{2^6}{2^{12}}</math> due to binomial theorem. | ||
+ | By symmetry, we must multiply by <math>2</math> to account for switching <math>R</math> and <math>S</math>, and another <math>2</math> for the case where <math>S=R=6</math>. Finally, we subtract the four overlapping cases <math>(0, 0), (0, 6), (6,0),</math> and <math>(6, 6)</math>, where each pair has format <math>(S, R)</math>. | ||
+ | Thus, the total probability is <math>\frac{4 \cdot 2^6 - 4}{2^{12}}</math>. | ||
+ | |||
+ | Now, there are <math>4</math> more pairs <math>(S,R)</math> that satisfy the condition, namely <math>(2, 3), (3, 4), (3, 2),</math> and <math>(4, 3)</math>. | ||
+ | |||
+ | Again, make the trivial observation that each of these have the same probability due to <math>R/S</math> invariance and similar distribution. | ||
+ | |||
+ | For a single pair, take <math>(2, 3)</math> as an example. We know <math>N=1</math>, and we fix <math>S</math>. The probability is therefore <math>\frac{\binom{6}{3} \binom{3}{1} \binom{3}{1}}{2^{12}} = \frac{180}{2^{12}}</math>. | ||
+ | |||
+ | Adding all of them up, we get <math>\frac{4 \cdot 2^6 - 4 + 180 \cdot 4}{2^{12}} = \frac{2^6+179}{2^{10}} = \frac{243}{1024}</math>. | ||
+ | Thus, the answer is <math>\boxed{A}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ~Apollo Luo, edited by Mathemagician108 | ||
==See also== | ==See also== | ||
− | {{AMC12 box|year=2024|ab=B|num-b=24| | + | {{AMC12 box|year=2024|ab=B|num-b=24|after=Last Problem}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:13, 16 November 2024
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
.
~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.