/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 On a multiple-choice exam with t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 12

On a multiple-choice exam with three possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing?

Short Answer

Expert verified
The probability of a student guessing and getting at least 4 correct answers is \(11/243\), or approximately 0.0452675.

Step by step solution

01

Understanding the binomial probability formula

The probability of getting exactly k successes in n independent Bernoulli trials with probability p of success on each trial can be calculated using the binomial probability formula: P(X = k) = \(\binom{n}{k}\) p^k (1-p)^(n-k) In this case, each question on the exam represents an independent Bernoulli trial with success probability p = 1/3 (since there are three possible answers, and only one correct answer). We are looking for the probability of getting at least four correct answers, so we will need to find P(X = 4) + P(X = 5).
02

Calculating the binomial coefficients

We will first need to compute the binomial coefficients \(\binom{n}{k}\) for the cases X = 4 and X = 5. These coefficients represent the number of ways we can choose k successes from n trials: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) For X = 4: \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\) For X = 5: \(\binom{5}{5} = \frac{5!}{5!(5-5)!} = 1\)
03

Computing the individual probabilities

Now, we will compute the probability of 4 and 5 correct answers using the binomial probability formula from Step 1: For X = 4: P(X = 4) = \(\binom{5}{4}\) (1/3)^4 (2/3)^1 = 5 * (1/81) * (2/3) = 10/243 For X = 5: P(X = 5) = \(\binom{5}{5}\) (1/3)^5 (2/3)^0 = 1 * (1/243) * 1 = 1/243
04

Computing the total probability

Finally, we can add the probabilities of getting 4 or 5 correct answers just by guessing: P(X >= 4) = P(X = 4) + P(X = 5) = 10/243 + 1/243 = 11/243 So, the probability of guessing and getting at least 4 correct answers is 11/243, or approximately 0.0452675.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that \(E\left[X^{2}\right] \geqslant(E[X])^{2}\). When do we have equality?

Calculate the moment generating function of the uniform distribution on \((0,1) .\) Obtain \(E[X]\) and \(\operatorname{Var}[X]\) by differentiating.

If the density function of \(X\) equals $$ f(x)=\left\\{\begin{array}{ll} c e^{-2 x}, & 02\\} ?\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent random variables, each having a uniform distribution over \((0,1)\). Let \(M=\operatorname{maximum}\left(X_{1}, X_{2}, \ldots, X_{n}\right) .\) Show that the distribution function of \(M, F_{M}(\cdot)\), is given by $$ F_{M}(x)=x^{n}, \quad 0 \leqslant x \leqslant 1 $$ What is the probability density function of \(M ?\)

The random variable \(X\) has the following probability mass function $$ p(1)=\frac{1}{2}, \quad p(2)=\frac{1}{3}, \quad p(24)=\frac{1}{6} $$ Calculate \(E[X]\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.