Difference between revisions of "2017 AMC 10B Problems/Problem 9"
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Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>1/3 * 1/3 * 2/3</math> of the time. Thus, <math>2/27 * 3</math> = <math>6/27</math>. | Case 2: They guess only two right. We pick one of the questions to get wrong, <math>3</math>, and this can happen <math>1/3 * 1/3 * 2/3</math> of the time. Thus, <math>2/27 * 3</math> = <math>6/27</math>. | ||
| − | So in total the two cases | + | So, in total the two cases combined equals <math>1/27 + 6/27</math> = <math>\boxed{\textbf{(D)}\ 7/27}</math>. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2017|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:29, 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. 
