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](https://artofproblemsolving.com/wiki/index.php/Probability) that the top face has an odd number of dots?
$(\mathrm {A}) \ \frac{5}{11} \qquad (\mathrm {B}) \ \frac{10}{21} \qquad (\mathrm {C})\ \frac{1}{2} \qquad (\mathrm {D}) \ \frac{11}{21} \qquad (\mathrm {E})\ \frac{6}{11}$
Solution 1
There are $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots total. [Casework](https://artofproblemsolving.com/wiki/index.php/Casework):

The dot is removed from an even face. There is a $\frac{2+4+6}{21} = \frac{4}{7}$ chance of this happening. Then there are 4 odd faces, giving us a probability of $\frac 47 \cdot \frac 46 = \frac{8}{21}$.
The dot is removed from an odd face. There is a $\frac{1+3+5}{21} = \frac{3}{7}$ chance of this happening. Then there are 2 odd faces, giving us a probability of $\frac 37 \cdot \frac 26 = \frac{1}{7}$.
Thus the answer is $\frac 8{21} + \frac 17 = \frac{11}{21} \Longrightarrow \mathrm{(D)}$.

Solution 2 (Alcumus)
The dot is chosen from the face with $n$ dots with probability $\frac{n}{21}$. Thus the face that originally has $n$ dots is left with an odd number of dots with probability $\frac{n}{21}$ if $n$ is even and $1 - n/21$ if $n$ is odd. Each face is the top face with probability $\frac{1}{6}$. Therefore the top face has an odd number of dots with probability\begin{align*}
&\frac{1}{6}\displaystyle\left(\displaystyle\left(1 - \frac{1}{21}\displaystyle\right) + \frac{2}{21} + \displaystyle\left(1 - \frac{3}{21}\displaystyle\right)
+ \frac{4}{21} + \displaystyle\left(1 - \frac{5}{21}\displaystyle\right) + \frac{6}{21}\displaystyle\right) \\
& \qquad = \frac{1}{6} \displaystyle\left(3 + \frac{3}{21}\displaystyle\right)\\
& \qquad = \frac{1}{6}\cdot \frac{66}{21} \\
& \qquad = \boxed{\frac{11}{21}}.
\end{align*}

Solution 3 (Alcumus)
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$\left(\frac{1}{3}\right)\left(\frac{1+3+5}{21}\right) +\left(\frac{2}{3}\right)\left(\frac{2+4+6}{21}\right) = \frac{33}{63} = \boxed{\frac{11}{21}}$.