AoPSWiki
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
Personal tools

2005 AMC 12A Problems/Problem 14

From AoPSWiki

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}) \ \f...

Solution

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)}.

See also

2005 AMC 12A (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Visit the AoPS Book Store.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us