Math Problem Statement

Solve this statistics problem

A student is given a test consisting of 19 multiple-choice questions, each having 5 possible answers. The student didn't study, so he decides to answer every question at random. A. For one question, what is the probability that a student will answer it correctly if he has 5 possible answers and just answers it at random? B. What is the probability that out of 19 questions, he answers no questions right? C. What is the probability that out of 19 questions, he answers exactly one question right?

Solution

Let's go through each part of the problem one by one.

Part A

For a single question with 5 possible answers, the probability of answering correctly at random is 15\frac{1}{5}.

So, the answer to part (A) is: 15\frac{1}{5}


Part B

For part (B), we want to find the probability that the student answers no questions correctly out of 19 questions.

Since each question has a 15\frac{1}{5} probability of being correct, the probability of getting a question wrong is 115=451 - \frac{1}{5} = \frac{4}{5}.

Let XX be the number of questions answered correctly. XX follows a Binomial distribution with parameters n=19n = 19 and p=15p = \frac{1}{5}.

The probability of answering no questions correctly (i.e., X=0X = 0) is: P(X=0)=(45)19P(X = 0) = \left(\frac{4}{5}\right)^{19}

Calculating this value: (45)190.0132\left(\frac{4}{5}\right)^{19} \approx 0.0132

So, the answer to part (B), rounded to four decimal places, is: 0.01320.0132


Part C

For part (C), we need to find the probability that the student answers exactly one question correctly out of 19.

Again, using the Binomial distribution with n=19n = 19 and p=15p = \frac{1}{5}, the probability of answering exactly one question correctly (i.e., X=1X = 1) is given by: P(X=1)=(191)(15)1(45)18P(X = 1) = \binom{19}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{18}

Calculating each component:

  • (191)=19\binom{19}{1} = 19
  • (15)1=0.2\left(\frac{1}{5}\right)^1 = 0.2
  • (45)180.0174\left(\frac{4}{5}\right)^{18} \approx 0.0174

Putting it all together: P(X=1)=19×0.2×0.01740.0661P(X = 1) = 19 \times 0.2 \times 0.0174 \approx 0.0661

So, the answer to part (C), rounded to four decimal places, is: 0.06610.0661


Summary of Answers

  • Part A: 15\frac{1}{5}
  • Part B: 0.01320.0132
  • Part C: 0.06610.0661

Would you like further explanations on any part of this problem, or have any specific questions?


Related Questions

  1. What would the probability be if there were 25 questions instead of 19?
  2. How does the probability of no correct answers change if each question has only 4 possible answers?
  3. What is the expected number of correct answers for this test?
  4. How do we calculate the standard deviation of correct answers in this scenario?
  5. What is the probability that the student answers at least one question correctly?

Tip

When dealing with probabilities in multiple-choice questions, using the Binomial distribution is very useful for calculating scenarios like "exactly" or "at least" a certain number of correct answers.

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Math Problem Analysis

Mathematical Concepts

Probability
Binomial Distribution

Formulas

Probability of a single correct answer = 1/5
Binomial probability formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Theorems

Binomial Distribution

Suitable Grade Level

Grades 9-12