2017 AMC 10B Problems/Problem 9
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 |
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