2017 AMC 10B Problems/Problem 9
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?
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 by TheBeautyofMath
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