/*! 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 3 A pair of standard dice are roll... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 3

A pair of standard dice are rolled. What is the probability of observing the following: a. The sum of the dice is equal to 7 b. The sum of the dice is equal to 9 c. The sum of the dice is less than or equal to 7

Short Answer

Expert verified
a. \(P(sum = 7) = \frac{1}{6}\) b. \(P(sum = 9) = \frac{1}{9}\) c. \(P(sum \leq 7) = \frac{7}{12}\)

Step by step solution

01

Part a: Probability of sum equal to 7

To find the probability of the sum of the dice equaling 7, we need to determine how many outcomes result in a sum of 7 and divide that by the total number of possible outcomes. The possible combinations that result in a sum of 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). There are 6 favorable outcomes, so the probability is: \(P(sum = 7) = \frac{6}{36} = \frac{1}{6} \)
02

Part b: Probability of sum equal to 9

To find the probability of the sum of the dice equaling 9, we need to determine how many outcomes result in a sum of 9 and divide that by the total number of possible outcomes. The possible combinations that result in a sum of 9 are: (3, 6), (4, 5), (5, 4), and (6, 3). There are 4 favorable outcomes, so the probability is: \(P(sum = 9) = \frac{4}{36} = \frac{1}{9} \)
03

Part c: Probability of sum less than or equal to 7

To find the probability of the sum of the dice being less than or equal to 7, we need to determine the total number of combinations that result in a sum of 1 through 7, and divide that by the total number of possible outcomes. We can count the favorable outcomes for each sum: - Sum of 2: (1,1) - 1 outcome - Sum of 3: (1,2), (2,1) - 2 outcomes - Sum of 4: (1,3), (2,2), (3,1) - 3 outcomes - Sum of 5: (1,4), (2,3), (3,2), (4,1) - 4 outcomes - Sum of 6: (1,5), (2,4), (3,3), (4,2), (5,1) - 5 outcomes - Sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - 6 outcomes In total, there are 1 + 2 + 3 + 4 + 5 + 6 = 21 favorable outcomes. So, the probability is: \(P(sum \leq 7) = \frac{21}{36} = \frac{7}{12} \)

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.

Standard Dice
Standard dice are the kind of dice commonly found in board games. They are six-sided cubes where each face shows a different number, ranging from 1 to 6. Each die is designed with perfectly even and balanced dimensions, which helps to ensure fair play.
When you roll a pair of standard dice, the outcome is determined by the numbers that appear on the top faces of both dice. The rolling dice can yield any combination between 1 to 6 on each die, leading to a fun world of possibilities.
Standard dice games rely on the fairness and equal probability of each face to show up, which means each individual face of the die has a chance of 1 out of 6 to appear on any given roll.
Outcomes
In probability, an outcome refers to a possible result of a random experiment. When rolling two standard dice, each die has 6 faces that result in a total of 36 possible outcomes.
  • Each outcome is unique and corresponds to a pair like (1, 1) meaning both dice show 1, or (2, 3) meaning one die shows 2 and the other shows 3.
  • This total of 36 outcomes arises from multiplying the number of sides on one die (6) by the number of sides on the second die (6).
Identifying each possible outcome is vital in calculating probabilities, as it provides the denominator in probability formulas.
Favorable Outcomes
Favorable outcomes are the specific results that meet the criteria of the event you are examining. For probability calculations, it's essential to count these outcomes accurately.
For instance, if you wish to find the probability of rolling a sum of 7 with a pair of dice, you identify which dice results meet this criterion. In the example, outcomes like (1, 6) or (4, 3) are favorable as they result in 7.
In any probability scenario, pinpointing these favorable outcomes is the first step that leads you to calculate the probability itself.
Probability Calculation
Probability calculation involves determining the likelihood of a particular event happening out of all possible outcomes. This is represented as a fraction where:
1. **The numerator** is the number of favorable outcomes.
2. **The denominator** is the total number of possible outcomes.
  • For example, the probability of rolling a sum of 7 with two dice is calculated as: \[ P( ext{sum} = 7) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6} \]
  • In a similar way, other scenarios can be calculated using the same formula and substituting the correct numbers for favorable and total outcomes.
Understanding probability calculation allows you to predict and quantify the chance of specific occurrences happening when performing an experiment in a probabilistic context.

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

Consider the probability distribution for molecular velocities in one dimension \(\left(v_{x}\right)\) given by \(P\left(v_{x}\right) d v_{x}=C e^{-m v^{2} x / 2 k T} d v_{x}\). a. Determine the normalization constant \(C\). b. Determine \(\left\langle v_{x}\right\rangle\) c. Determine \(\left\langle v_{x}^{2}\right\rangle\) d. Determine the variance.

Radio station call letters consist of four letters (for example, KUOW ). a. How many different station call letters are possible using the 26 letters in the English alphabet? b. Stations west of the Mississippi River must use the letter \(\mathrm{K}\) as the first call letter. Given this requirement, how many different station call letters are possible if repetition is allowed for any of the remaining letters? c. How many different station call letters are possible if repetition is not allowed for any of the letters?

Imagine performing the coin-flip experiment of Problem \(\mathrm{P} 29.19\), but instead of using a fair coin, a weighted coin is employed for which the probability of landing heads is two-fold greater than landing tails. After tossing the coin 10 times, what is the probability of observing the following specific outcomes: a. no heads b. two heads c. five heads d. eight heads

a. Consider assigning phone numbers to an area where each number consists of seven digits, each of which can be between 0 and \(9 .\) How many phone numbers can be assigned to the area? b. To serve multiple areas, you decide to introduce an area code that serves as an identifier. In the United States, area codes were initially three-digit numbers in which the first digit ranged from 2 to \(9,\) the second digit was either 0 or 1 and the third digit ranged from 1 to \(9 .\) How many areas could be specified using this system? c. Using area codes with individual phone numbers, how many unique phones could this system support?

Evaluate the following: a. the number of permutations employing all objects in a six-object set b. the number of permutations employing four objects from a six-object set c. the number of permutations employing no objects from a six-object set d. \(P(50,10)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

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.