Difference between revisions of "2017 AMC 10B Problems/Problem 9"
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The probability of getting each question incorrect is <math>\frac{2}{3}</math>. Thus, the probability of getting all questions incorrect is <math>\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}</math>. | The probability of getting each question incorrect is <math>\frac{2}{3}</math>. Thus, the probability of getting all questions incorrect is <math>\frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{8}{27}</math>. | ||
| − | Case 2: The contestant gets one question right. There are 3 ways the contestant can get one question correct since there are 3 questions. The probability of guessing correctly is <math>\frac{1}{3}</math> so the probability of guessing one correctly and two incorrectly is <math> | + | Case 2: The contestant gets one question right. There are 3 ways the contestant can get one question correct since there are 3 questions. The probability of guessing correctly is <math>\frac{1}{3}</math> so the probability of guessing one correctly and two incorrectly is <math> \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{9}</math>. Each of these 3 ways has a probability associated with it, so probability of getting 1 correct is <math>3 * \frac{2}{3}</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 so 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 so the answer is <math>1-\frac{20}{27} = \boxed{\textbf{(D)}\frac{7}{27}}</math> | ||
Revision as of 11:58, 17 August 2021
Contents
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 getting each question incorrect is 
. Thus, the probability of getting all questions incorrect is 
.
Case 2: The contestant gets one question right. There are 3 ways the contestant can get one question correct since there are 3 questions. The probability of guessing correctly is 
so the probability of guessing one correctly and two incorrectly is 
. Each of these 3 ways has a probability associated with it, so probability of getting 1 correct is 
The sum of the two cases is 
. This is the complement of what we want so the answer is 
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/XRfOULUmWbY?t=482
~IceMatrix
Video Solution
https://youtu.be/IRyWOZQMTV8?t=1029
~ pi_is_3.14
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. 
