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2025 AMC 10A Problems/Problem 16

Question

There are three jars. Each of three coins is placed in one of the tree jars, chosen at random and independently of the placement of the other coins. What is the expected number of coins in a jar with the most coins?

$\textbf{(A) } \frac{4}{3}\qquad\textbf{(B) } \frac{13}{9}\qquad\textbf{(C) } \frac{5}{8}\qquad\textbf{(D) } \frac{17}{9}\qquad\textbf{(E) } 2$

Solution

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 coin is alone in 3 ways. Therefore, there are $3 \cdot 2 \cdot 3 = 18$ 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

-Boywithnuke(Goal to 10 followers)

~ Minor edits by SixthGradeBookWorm927

Video Solution (In 1 Min)

https://youtu.be/8iDugBBzei8?si=QXHiwxmA5eqXyEZY ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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