Difference between revisions of "2014 CEMC Gauss (Grade 7) Problems/Problem 4"
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The spinner shown is divided into 6 sections of equal size. | The spinner shown is divided into 6 sections of equal size. | ||
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| − | What is the probability of landing on a section that contains the letter P using this spinner | + | What is the probability of landing on a section that contains the letter P using this spinner? |
<math> \textbf{(A)}\ \frac{3}{6} \qquad\textbf{(B)}\ \frac{4}{6} \qquad\textbf{(C)}\ \frac{5}{6} \qquad\textbf{(D)}\ \frac{2}{6} \qquad\textbf{(E)}\ \frac{1}{6} </math> | <math> \textbf{(A)}\ \frac{3}{6} \qquad\textbf{(B)}\ \frac{4}{6} \qquad\textbf{(C)}\ \frac{5}{6} \qquad\textbf{(D)}\ \frac{2}{6} \qquad\textbf{(E)}\ \frac{1}{6} </math> | ||
==Solution== | ==Solution== | ||
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Thus, the probability is <math>\boxed {{(D) } \frac{2}{6}}</math>. | Thus, the probability is <math>\boxed {{(D) } \frac{2}{6}}</math>. | ||
| + | {{CEMC box|year=2014|competition=Gauss (Grade 7)|num-b=3|num-a=5}} | ||
Latest revision as of 11:55, 18 October 2025
Problem
The spinner shown is divided into 6 sections of equal size.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
What is the probability of landing on a section that contains the letter P using this spinner?
Solution
To find the probability of landing on a section with the letter P, we can divide the number of sections that contain P by the total number of sections.
As stated in the problem, there are 
sections of equal size. Looking at the spinner, we can notice that P appears 
times in the spinner.
Thus, the probability is 
.
| 2014 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| CEMC Gauss (Grade 7) | ||
