Difference between revisions of "2017 AMC 10B Problems/Problem 9"
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The sum of the two cases is <math>\frac{8}{27} + \frac{4}{9} = \frac{20}{27}</math>. This is the complement of what we want to the answer is <math>1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}</math> | The sum of the two cases is <math>\frac{8}{27} + \frac{4}{9} = \frac{20}{27}</math>. This is the complement of what we want to the answer is <math>1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}</math> | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/XYeexmAyVzQ | ||
| + | |||
| + | ~savannahsolver | ||
==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 15:46, 22 July 2020
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 1
There are two ways the contestant can win.
Case 1: The contestant guesses all three right. This can only happen 
of the time.
Case 2: The contestant guesses 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 
= 
.
Solution 2 (complementary counting)
Complementary counting is good for solving the problem and checking work if you solved it using the method above.
There are two ways the contestant can lose.
Case 1: The contestant guesses zero questions correctly.
The probability of guessing incorrectly for each question is 
. Thus, the probability of guessing all questions incorrectly is 
.
Case 2: The contestant guesses one question correctly. There are 3 ways the contestant can guess one question correctly since there are 3 questions. The probability of guessing correctly is 
so the probability of guessing one correctly and two incorrectly is 
.
The sum of the two cases is 
. This is the complement of what we want to the answer is 
Video Solution
~savannahsolver
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. 
