/*! 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 8 What is the expected sum of the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 8

What is the expected sum of the numbers that appear when three fair dice are rolled?

Short Answer

Expert verified
10.5

Step by step solution

01

Determine the expected value for one die

Each side of a fair six-sided die (1, 2, 3, 4, 5, 6) has an equal probability of \(\frac{1}{6}\). The expected value (E) of a die can be calculated using the formula for the expected value of a uniform distribution:\[ E = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \]
02

Calculate the expected sum for three dice

The expected sum when multiple dice are rolled is the sum of the expected values of each individual die. Since each die has an expected value of 3.5, and there are three dice, the calculation is:\[ E(\text{sum}) = 3.5 + 3.5 + 3.5 = 3.5 \times 3 = 10.5 \]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a way to measure the likelihood of a certain event happening. It ranges from 0 to 1, where 0 means the event cannot happen, and 1 means the event will certainly happen. In the context of rolling a die, each side has an equal chance of appearing. Since a fair six-sided die has six faces, each face has a probability of \( \frac{1}{6} \). When we roll three dice, we consider the probability for each of the dice to contribute to the result independently.
Expected Value
The expected value (E) is a fundamental concept in probability and statistics. It represents the average outcome of a random event if the same experiment is repeated many times. For a single six-sided die, the expected value is calculated by adding the values of all possible outcomes and dividing by the number of outcomes: \[ E = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \]. This means that, on average, you can expect a roll of a die to result in 3.5 over a large number of trials.
Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely. In the case of a fair six-sided die, each number from 1 to 6 is equally likely to occur, representing a uniform distribution. This is why each number has a probability of \( \frac{1}{6} \). This equal likelihood simplifies the calculation of the expected value and ensures that each face contributes equally to the average outcome.
Sum of Random Variables
When dealing with multiple dice, the sum of their outcomes is a random variable in itself. The expected value of the sum of independent random variables is the sum of their expected values. Therefore, if each die has an expected value of 3.5, the expected sum for three dice is: \[ E(\text{sum}) = 3.5 + 3.5 + 3.5 = 10.5 \]. This rule holds generally and is extremely useful in various applications of probability and statistics, making calculations straightforward when dealing with sums of independent events.

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

Use the law of total expectation to find the average weight of a breeding elephant seal, given that 12% of the breeding elephant seals are male and the rest are female, and the expected weights of a breeding elephant seal is 4200 pounds for a male and 1100 pounds for a female.

What is the expected number of times a 6 appears when a fair die is rolled 10 times?

What is the conditional probability that a randomly generated bit string of length four contains at least two consecutive 0s, given that the first bit is a 1? (Assume the probabilities of a 0 and a 1 are the same.)

Prove Theorem \(2,\) the extended form of Bayes' theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \(F_{1}, F_{2}, \ldots, F_{n}\) are mutually exclusive events such that \(\bigcup_{i=1}^{n} F_{i}=S .\) Assume that \(p(E) \neq 0\) and \(p\left(F_{i}\right) \neq 0\) for \(i=1,2, \ldots, n .\) Show that $$ p\left(F_{j} | E\right)=\frac{p\left(E | F_{j}\right) p\left(F_{j}\right)}{\sum_{i=1}^{n} p\left(E | F_{i}\right) p\left(F_{i}\right)} $$ \(\left[\text {Hint} : \text { Use the fact that } E=\bigcup_{i=1}^{n}\left(E \cap F_{i}\right) .\right]\)

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a) a 0 bit and a 1 bit are equally likely. b) the probability that a bit is a 1 is 0.6 . c) the probability that the ith bit is a 1 is 1\(/ 2^{i}\) for \(i=\) \(\quad 1,2,3, \ldots, 10\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

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.