Question

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

$\dfrac{4}{3}$

$\dfrac{39}{27}$

$\dfrac{5}{3}$

$\dfrac{17}{9}$

$2$

Answer: D

Concepts:expected value casework

Difficulty rating: 1630

Solution:

There are $3^3 = 27$ equally likely placements. Of these, $3$ pile all coins into one jar (max $3$ ), and $6$ put one coin in each jar (max $1$ ). The other $18$ split $2$ - $1$ (max $2$ ). So the expected maximum is $\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{51}{27} = \frac{17}{9}.$ Therefore, the answer is D.

2025 AMC 10A Problem 16

Answer: D

Difficulty rating: 1630

Solution:

Problem 16 in Other Years