Difference between revisions of "2017 AMC 10B Problems/Problem 9"
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<math>\textbf{(A)}\ \frac{1}{27}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{7}{27}\qquad\textbf{(E)}\ \frac{1}{2}</math> | <math>\textbf{(A)}\ \frac{1}{27}\qquad\textbf{(B)}\ \frac{1}{9}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{7}{27}\qquad\textbf{(E)}\ \frac{1}{2}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | There are two ways the contestant can win. | |
+ | |||
+ | Case 1: The contestant guesses 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: The contestant guesses 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} * 3</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>. | ||
+ | |||
+ | |||
+ | More detailed explanation: For case 1, the contestant must guess all three correctly. The probability of guessing one problem right is <math>\frac{1}{3}</math>, so the probability of getting all three right is <math>\left(\frac{1}{3}\right)^{3}</math>. | ||
+ | For case 2: we must choose one of the problems to answer correctly and two to answer incorrectly. The probabilities for guessing correctly and incorrectly are <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math>, respectively. So we have <math>\left(\frac{1}{3}\right)^{2}\cdot\frac{2}{3}\cdot3</math>. The answer is the sum of probabilities of case 1 and 2, since there are no overcounts. <math>\frac{1}{27}+\frac{6}{27}=\frac{7}{27}</math>. | ||
+ | |||
+ | ==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 <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> \frac{1}{3} * \frac{2}{3} * \frac{2}{3} = \frac{4}{27}</math>. Each of these 3 ways has a probability associated with it, so probability of getting 1 correct is <math>3 * \frac{4}{27} = \frac{4}{9} </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> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/XYeexmAyVzQ | ||
+ | |||
+ | ~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== | ==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}} |
Latest revision as of 08:16, 4 November 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 =
.
More detailed explanation: For case 1, the contestant must guess all three correctly. The probability of guessing one problem right is , so the probability of getting all three right is
.
For case 2: we must choose one of the problems to answer correctly and two to answer incorrectly. The probabilities for guessing correctly and incorrectly are
and
, respectively. So we have
. The answer is the sum of probabilities of case 1 and 2, since there are no overcounts.
.
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.