Difference between revisions of "2017 AMC 10B Problems/Problem 9"
m (→Video Solution 2) |
Pi is 3.14 (talk | contribs) (→Video Solution by TheBeautyofMath) |
||
Line 36: | Line 36: | ||
~IceMatrix | ~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}} |
Revision as of 21:51, 17 January 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 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
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.