Problem:
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?
Answer Choices:
A. 115​
B. 2110​
C. 21​
D. 2111​
E. 116​
Solution:
A standard die has a total of 21 dots. For 1≤n≤6, a dot is removed from the face with n dots with probability n/21. Thus the face that originally has n dots is left with an odd number of dots with probability n/21 if n is even and 1−n/21 if n is odd. Each face is the top face with probability 1/6. Therefore the top face has an odd number of dots with probability
61​((1−211​)+212​+(1−213​)+214​+(1−215​)+216​)=61​(3+213​)=61​⋅2166​=2111​​​
\section*{OR}
The probability that the top face is odd is 1/3 if a dot is removed from an odd face, and the probability that the top face is odd is 2/3 if a dot is removed from an even face. Because each dot has the probability 1/21 of being removed, the top face is odd with probability
(31​)(211+3+5​)+(32​)(212+4+6​)=6333​=2111​​
The problems on this page are the property of the MAA's American Mathematics Competitions