What is the possibility that a student will answer at least three questions correctly by random guessing on an exam with 10 multiple-choice questions each of which has four options? How would I figure this out?
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Anonymous
ANSWERS: 1
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Each choice has four possibilities. So the total number of possibilities of answers is 4^10. To get the "three or fewer questions right" probability( which we can then subtract from 1), we need to add together the number of ways to get 0,1,2 and 3 questions right. How many ways are there to get 0 questions right? 3^10. Why? For each question, three answers are wrong. How many ways are there to get 1 questions right? 1 * 3^9. Multiply by 10, since there are 10 ways (10 questions) to have one answer right. How many ways are there to get 2 questions right? 1 * 1* 3^8. Multiply by 45, since there are 45 ways to have two answers right. And so on.
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