Difference between revisions of "2017 AMC 10B Problems/Problem 9"
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Case 1: They guess all three right. This can only happen <math>\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}</math> of the time. | Case 1: They guess all three right. This can only happen <math>\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}</math> of the time. | ||
− | Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>\frac{1}{3} * \frac{1}{3} * \frac{2} | + | Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>\frac{1}{3} * \frac{1}{3} * \frac{2}{3}</math> of the time. Thus, <math>\frac{2}{27} * \frac{3}{1}</math> = <math>\frac{6}{27}</math>. |
So, in total the two cases combined equals <math>\frac{1}{27} + \frac{6}{27}</math> = <math>\boxed{\textbf{(D)}\ \frac{7}{27}}</math>. | So, in total the two cases combined equals <math>\frac{1}{27} + \frac{6}{27}</math> = <math>\boxed{\textbf{(D)}\ \frac{7}{27}}</math>. |
Revision as of 15:19, 16 February 2017
Problem
A radio program has a quiz consisting of multiple-choice questions, each with choices. A contestant wins if he or she gets or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?
Solution
There are two ways that the contestant can win.
Case 1: They guess all three right. This can only happen of the time.
Case 2: They guess only two right. We pick one of the questions to get wrong, , and this can happen of the time. Thus, = .
So, in total the two cases combined equals = .
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.