2005 AMC 12A Problems/Problem 14

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?

Solutions

Solution 1

There are dots total. Casework:

• The dot is removed from an even face. There is a chance of this happening. Then there are 4 odd faces, giving us a probability of .
• The dot is removed from an odd face. There is a chance of this happening. Then there are 2 odd faces, giving us a probability of .

Thus the answer is .

Solution 2 (Alcumus)

The dot is chosen from the face with dots with probability . Thus the face that originally has dots is left with an odd number of dots with probability if is even and if is odd. Each face is the top face with probability . 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 if a dot is removed from an odd face, and the probability that the top face is odd is if a dot is removed from an even face. Because each dot has the probability of being removed, the top face is odd with probability\[ \displaystyle\left(\frac{1}{3}\displaystyle\right)\displaystyle\left(\frac{1+3+5}{21}\displaystyle\right) + \displaystyle\left(\frac{2}{3}\displaystyle\right)\displaystyle\left(\frac{2+4+6}{21}\displaystyle\right) = \frac{33}{63} = \boxed{\frac{11}{21}}. \]

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.