Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.
Note: My solution does not provide an exact probability, but is a good estimate, which is how this problem was designed. Most of the solutions on this page will give a decent estimate. This is due to the fact that the binomial expansions are symmetric for even vs odd. Try experimenting to determine why this is true.
~lprado
~AtharvNaphade
~eevee9406
~Teddybear0629
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?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$