/*! 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 6 A fair die is rolled 10 times. C... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 6

A fair die is rolled 10 times. Calculate the expected sum of the 10 rolls.

Short Answer

Expert verified
The expected sum of 10 rolls of a fair die is 35.

Step by step solution

01

Determine the expected value of a single roll

The expected value of a single roll is the average value of all possible outcomes. In the case of a fair six-sided die, the possible outcomes are the numbers 1 through 6, and the probability of each outcome is 1/6. The expected value can be calculated using the following formula: Expected Value = 危 (value 脳 probability) For a fair die, the expected value is: Expected Value = \(1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6})\)
02

Calculate the expected value of a single roll

Using the above formula, we can now calculate the expected value of a single roll: Expected Value = \(1(\frac{1}{6}) + 2(\frac{1}{6}) + 3(\frac{1}{6}) + 4(\frac{1}{6}) + 5(\frac{1}{6}) + 6(\frac{1}{6})\) Expected Value = \(\frac{1}{6} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6}\) Expected Value = \(\frac{21}{6}\) Expected Value = 3.5
03

Calculate the expected sum of 10 rolls

Now that we know the expected value of a single roll (3.5), we can multiply it by the number of rolls (10) to find the expected sum of the 10 rolls: Expected Sum = (Expected Value of a Single Roll) 脳 (Number of Rolls) Expected Sum = 3.5 脳 10 = 35 So, the expected sum of 10 rolls of a fair die is 35.

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Key Concepts

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

Probability
The concept of probability revolves around the likelihood of an event occurring. When rolling a fair die, each face has an equal chance of landing face up. This is what we call uniform probability. With a six-sided die, each number from 1 to 6 has a probability of 1/6 of appearing on any given roll.

Probability forms the backbone of understanding expected values. It's essentially the assessment of all possible outcomes and how likely each is to occur.
  • If an experiment has multiple outcomes, the probability of each outcome must add up to 1.
  • For example, the sum of probabilities for a die鈥檚 outcomes 1, 2, 3, 4, 5, and 6 is exactly 1, confirming our calculations are correct.
  • Understanding probability helps us make predictions about future events, based on the likelihood of various outcomes.
Random Variables
A random variable is a numerical representation of the outcomes in a probability experiment. In the context of rolling a die, each face represents a different numerical outcome. We treat these possible outcomes as random variables, since they're based on chance.

The concept of random variables helps us quantify the expected value in the exercise.
  • Random variables can be categorized as discrete or continuous. A die outcome is a discrete random variable because it has specific, separate values: 1 to 6.
  • In the case of our die roll, the values we assign depend solely on probability, where each face of the die represents a potential value the random variable can take.
  • The sum of random variables, such as total points from multiple rolls, illustrates how individual random outcomes aggregate to form a larger picture.
Summation
Summation is the process of adding a sequence of numbers, and it is denoted by the symbol \(\Sigma\). In our exercise, summation helps calculate the expected value of rolling a die multiple times.

Summation encompasses the entire set of outcomes of a die roll, allowing us to find the average outcome, also known as the expected value.
  • The calculation of an expected value involves multiplying each outcome by its probability and then summing all products together.
  • For a single die roll, this is shown in the formula: \(1(\frac{1}{6}) + 2(\frac{1}{6}) +...+ 6(\frac{1}{6}) = \frac{21}{6} = 3.5\).
  • When rolling the die 10 times, the summation concept simplifies to multiplying the single roll expected value by the number of rolls, making the computation straightforward: \(3.5 \times 10 = 35\).
Integration of summation with expected values allows us to generalize results over numerous trials, aiding in understanding long-term outcomes of probabilistic events.

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Most popular questions from this chapter

The county hospital is located at the center of a square whose sides are 3 miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are \((0,0),\) to the point \((x, y)\) is \(|x|+|y| .\) If an accident occurs at a point that is uniformly distributed in the square, find the expected travel distance of the ambulance.

Between two distinct methods for manufacturing certain goods, the quality of goods produced by method \(i\) is a continuous random variable having distribution \(F_{i}, i=\) 1,2. Suppose that \(n\) goods are produced by method 1 and \(m\) by method \(2 .\) Rank the \(n+m\) goods according to quality, and let$$X_{j}=\left\\{\begin{array}{ll}1 & \text { if the } j \text { th best was produced from } \\\& \text { method } 1 \\\2 & \text { otherwise }\end{array}\right.$$ For the vector \(X_{1}, X_{2}, \ldots, X_{n+m},\) which consists of \(n\) 1's and \(m\) 2's, let \(R\) denote the number of runs of \(1 .\) For instance, if \(n=5, m=2,\) and \(X=1,2,1,1,1,1,2,\) then \(R=2 .\) If \(F_{1}=F_{2}\) (that is, if the two methods produce identically distributed goods), what are the mean and variance of \(R ?\)

Consider 3 trials, each having the same probability of success. Let \(X\) denote the total number of successes in these trials. If \(E[X]=1.8,\) what is (a) the largest possible value of \(P\\{X=3\\} ?\) (b) the smallest possible value of \(P\\{X=3\\} ?\) In both cases, construct a probability scenario that results in \(P\\{X=3\\}\) having the stated value.

Let \(X_{1}, \ldots\) be independent random variables with the common distribution function \(F,\) and suppose they are independent of \(N,\) a geometric random variable with parameter \(p .\) Let \(M=\max \left(X_{1}, \ldots, X_{N}\right)\) (a) Find \(P\\{M \leq x\\}\) by conditioning on \(N\) (b) Find \(P\\{M \leq x | N=1\\}\) (c) Find \(P\\{M \leq x | N>1\\}\) (d) Use (b) and (c) to rederive the probability you found in (a).

A group of \(n\) men and \(n\) women is lined up at random. (a) Find the expected number of men who have a woman next to them. (b) Repeat part (a), but now assuming that the group is randomly seated at a round table.
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