Difference between revisions of "2023 AMC 10B Problems/Problem 21"
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The last expression above is the number of ways to get all three bins with odd numbers of balls. | The last expression above is the number of ways to get all three bins with odd numbers of balls. | ||
Therefore, this happens with probability | Therefore, this happens with probability | ||
− | < | + | <cmath> |
\begin{align*} | \begin{align*} | ||
\frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}} | \frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}} | ||
− | & \approx \boxed{\textbf{(E) | + | & \approx \boxed{\textbf{(E) } \frac{1}{4}}. |
\end{align*} | \end{align*} | ||
− | + | </cmath> | |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 18:17, 15 November 2023
Contents
Problem
Each of 2023 balls is randomly placed into one of 3 bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
Solution 1
We first examine the possible arrangements for parity of number of balls in each box for balls.
If a denotes an even number and a
denotes an odd number, then the distribution of balls for
balls could be
or
. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From , it is not possible to get to all odd by adding one ball; we could either get
or
. For the other
cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have possible slot the ball can go into (for
, for example, the new ball must go in the center slot to make
) out of the
slots, so there's a
chance. We have a
chance of getting one of these working cases, so our answer is
~Technodoggo
Solution 2
We will start with all the balls outside of the boxes, and distribute them as follows:
We put balls into the first box. There is (obviously) a roughly
probability
is odd (It's okay to not use the exact probability since the problem asks for the closest answer choice, and the answer choices aren't very close to each other).
We put balls into the second box. There is also a roughly
probability
is odd.
If both and
are odd, then the number of balls which go into the third box must also be odd, since 2023 is odd.
Additionally,
and
clearly must both be odd in order for the problem conditions to be satisfied.
Therefore our answer is the probability both and
are odd, which is approximately
~kjljixx
Solution 3
We use the generating functions approach to solve this problem.
Define .
We have
First, we set ,
,
.
We get
Second, we set ,
,
.
We get
Third, we set ,
,
.
We get
Fourth, we set ,
,
.
We get
Taking , we get
The last expression above is the number of ways to get all three bins with odd numbers of balls.
Therefore, this happens with probability
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)