AMC10 2025 A

We have three coins and three jars. Each coin is placed independently and randomly into one of the jars. Let $M$ be the maximum number of coins in any jar. We want to compute the expected value of $M$. Step 1: Count total outcomes Each coin has $3$ choices, so the total number of equally likely placements is $3^3 = 27$. Step 2: Casework on the maximum number of coins Case 1: $M = 1$. This occurs when each jar has exactly one coin. There are $3! = 6$ assignments of coins to jars. Hence, $\Pr(M=1) = \frac{6}{27} = \frac{2}{9}$. Case 2: $M = 3$. This occurs when all three coins fall into the same jar. There are $3$ jars to choose from, so $\Pr(M=3) = \frac{3}{27} = \frac{1}{9}$. Case 3: $M = 2$. This occurs when one jar has $2$ coins, another jar has $1$ coin, and the last jar has $0$ coins. We can choose which jar gets $2$ coins in $3$ ways, which jar gets $1$ coin in $2$ ways, and which $2$ coins out of the $3$ go into the jar with two coins, so we multiply by $\dbinom{3}{2}$, which is just $3$ (note we don't have to do this for the earlier cases because for case $2$, all $3$ coins go into one jar, and for case $1$, the factorial already accounts for that). Therefore, there are $3^2 \cdot 2 = 18$ outcomes. Thus, $\Pr(M=2) = \frac{18}{27} = \frac{2}{3}$. Step 3: Compute the expected value The expected value of $M$ is $\mathbb{E}[M] = 1 \cdot \frac{2}{9} + 2 \cdot \frac{2}{3} + 3 \cdot \frac{1}{9}$. Converting everything to ninths, we have $\mathbb{E}[M] = \frac{2}{9} + \frac{12}{9} + \frac{3}{9} = \frac{17}{9}$. Hence, the expected number of coins in the jar with the most coins is$\boxed{\text{(D) }\frac{17}{9}}$ . $Pr$=Probability of $\mathbb{E}$=Expected value As described in the solution, there are $3^3=27$ ways of distributing the coins into the $3$ jars. Because there are $6$ ways for M=1 and $3$ ways for M=3, there are $27 - 6 - 3 = 18$ ways for $M=2$.