Difference between revisions of "2005 AMC 12A Problems/Problem 14"

Latest revision as of 16:44, 3 July 2024

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?

$(\mathrm {A}) \ \frac{5}{11} \qquad (\mathrm {B}) \ \frac{10}{21} \qquad (\mathrm {C})\ \frac{1}{2} \qquad (\mathrm {D}) \ \frac{11}{21} \qquad (\mathrm {E})\ \frac{6}{11}$

Solutions

Solution 1

There are $1 + 2 + 3 + 4 + 5 + 6 = 21$ dots total. 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 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 $\left(\frac{1}{3}\right)\left(\frac{1+3+5}{21}\right) +\left(\frac{2}{3}\right)\left(\frac{2+4+6}{21}\right) = \frac{33}{63} = \boxed{\frac{11}{21}}$ .

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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

@@ Line 16: / Line 16: @@
 === Solution 2 (Alcumus) ===
-The dot is chosen from the face with <math>n</math> dots with probability <math>\frac{n}{21}</math>. Thus the face that originally has <math>n</math> dots is left with an odd number of dots with probability <math>\frac{n}{21}</math> if <math>n</math> is even and <math>1 - n/21</math> if <math>n</math> is odd. Each face is the top face with probability <math>\frac{1}{6}</math>. 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 <math>1/3</math> if a dot is removed from an odd face, and the probability that the top face is odd is <math>2/3</math> if a dot is removed from an even face. Because each dot has the probability <math>1/21</math> of being removed, the top face is odd with probability<math>\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}}</math>.

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