Problem:
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?
Answer Choices:
A. 34
B. 913
C. 35
D. 917
E. 2
Solution:
We have 3 coins and 3 jars. Each coin is placed independently and uniformly at random into one of the 3 jars.
Hence, there are 33=27 possible outcomes, all equally likely.
Let M denote the number of coins in the jar with the most coins.
Each possible arrangement of the 3 coins among 3 jars must result in one of the following occupancy types:
Type(3,0,0)(2,1,0)(1,1,1)DescriptionAll 3 coins in one jarTwo in one jar, one in anotherOne in each jarMaximum occupancy321
We can also calculate the number of possible assignments of each type:
Type (3,0,0):Type (2,1,0):Type (1,1,1):3 ways (choose which jar gets all 3 coins).(23)(choose the 2 coins together)×3(choose jar for them)×2(choose jar for single coin)=3×3×2=18.3!=6 ways (each coin in a distinct jar).
We finish by calculating the expected value of M:
E[M]=273⋅3+18⋅2+6⋅1=279+36+6=2751=(D) 917.
The problems on this page are the property of the MAA's American Mathematics Competitions