Chapter 24: Problem 169
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is [2013] (A) \(\frac{13}{3^{5}}\) (B) \(\frac{11}{3^{5}}\) (C) \(\frac{10}{3^{5}}\) (D) \(\frac{17}{3^{5}}\)
Short Answer
Step by step solution
Understand the Problem
Identify the Probabilities
Use Binomial Probability Formula
Calculate the Probability of 4 Correct Answers
Calculate the Probability of 5 Correct Answers
Add Probabilities of Getting 4 or 5 Correct Answers
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Distribution
- \( n \): number of trials (in our case, the number of questions)
- \( p \): probability of success on a single trial (probability of guessing correctly)
Multiple Choice Examination
- Each question is independent of the others. Answering one question correctly doesn't affect the others.
- The probability of correctly guessing an answer amongst three choices is \( \frac{1}{3} \), since there is only one right answer.
- The examination is structured so that random guessing can lead to a certain distribution of correct and incorrect answers.
Binomial Probability Formula
- \( n \) is the number of trials (questions in the exam).
- \( k \) is the number of successes we want to find the probability for (correct answers).
- \( p \) is the probability of a success on any given trial.
- \( \binom{n}{k} \) represents the number of combinations of \( n \) trials taken \( k \) at a time.