/*! 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 23 A pair of fair dice are rolled. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 23

A pair of fair dice are rolled. What is the probability that the second die lands on a higher value than does the first?

Short Answer

Expert verified
The probability that the second die lands on a higher value than the first die is \(\frac{5}{12}\).

Step by step solution

01

Calculate the total possible outcomes

Since we are rolling two fair dice, each die can land on any of the numbers from 1 to 6. Therefore, we have 6 possible outcomes for each die. When we are rolling two dice together, the total possible outcomes can be calculated by multiplying the number of possible outcomes for each die (6 for each die, in this case). So the total possible outcomes are \(6 \times 6 = 36\).
02

Calculate the number of favorable outcomes

The favorable outcomes are the cases where the second die lands on a higher value than the first die. We can list the possible favorable outcomes as pairs (First die, Second die) and count them, or we can count them systematically by considering each possible value for the first die and counting how many favorable outcomes result from that value: - If the first die is 1, the second die can be any of 2, 3, 4, 5, or 6. So we have 5 favorable outcomes. - If the first die is 2, the second die can be any of 3, 4, 5, or 6. So we have 4 favorable outcomes. - If the first die is 3, the second die can be any of 4, 5, or 6. So we have 3 favorable outcomes. - If the first die is 4, the second die can be any of 5, or 6. So we have 2 favorable outcomes. - If the first die is 5, the second die can be 6. So we have 1 favorable outcome. - If the first die is 6, there are no favorable outcomes. So, the total number of favorable outcomes is \(5 + 4 + 3 + 2 + 1 = 15\).
03

Calculate the probability

To find the probability of the second die landing on a higher value than the first die, we'll divide the number of favorable outcomes by the total possible outcomes. Hence, the probability is given by: \(P(\text{Second die higher than first die}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{15}{36}\) We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor (3), and we'll get: \(P(\text{Second die higher than first die}) = \frac{15}{36} = \frac{5}{12}\) So, the probability that the second die lands on a higher value than the first die is \(\frac{5}{12}\).

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.

Fair Dice
When we discuss fair dice, we're talking about dice where each face has an equal chance of landing face up. Imagine a classic six-sided die鈥攜ou're equally likely to roll any number from 1 to 6. This uniform probability is crucial in games of chance, ensuring that no number is favored over others.

This concept of fairness is essential in probability calculations because it forms the basis of predictable outcomes. With fair dice, each roll is an independent event, meaning the result of one roll does not affect the outcomes of the next.
  • This independence ensures that each face has a consistent 1 in 6 chance of appearing each time you roll.
  • This consistency allows us to accurately predict theoretical probabilities when combining dice, like in our exercise where two dice are rolled.
Favorable Outcomes
In probability theory, favorable outcomes are those specific outcomes of an experiment that meet the criteria of our interest. For this exercise, these criteria involve the second die showing a higher number than the first.

We consider each possible number on the first die and count how many numbers the second die can display to satisfy our condition. For example, if the first die shows a 1, the favorable outcomes are when the second die shows 2, 3, 4, 5, or 6. As the number on the first die increases, the number of favorable outcomes decreases:
  • If the first die is 1, there are 5 favorable outcomes for the second die.
  • If the first die is 2, there are 4.
  • If the first die is 3, there are 3, and so on.


Counting all these favorable pairs, we obtain 15 favorable outcomes in total when combining two dice.
Probability Calculation
Calculating probability involves determining how likely it is for a certain event to occur. In our exercise, it's about calculating how likely it is that the second die will end up with a higher value than the first.

Probability is calculated using the formula: \[ P( ext{Event}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} \]In our case:
  • The number of favorable outcomes is 15.
  • The total possible outcomes when rolling two dice is 36.


This gives us a probability of \( \frac{15}{36} \), which simplifies to \( \frac{5}{12} \). This simplified fraction represents the chances of the second die landing on a higher number than the first, highlighting the importance of comprehending favorable outcomes and total possibilities in determining likelihood.

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

In a state lottery, a player must choose 8 of the numbers from 1 to 40 . The lottery commission then performs an experiment that selects 8 of these 40 . numbers. Assuming that the choice of the lottery commission is equally likely to be any of the \(\left(\begin{array}{c}40 \\ 8\end{array}\right)\) combinations, what is the probability that a player has (a) all 8 of the numbers selected; (b) 7 of the numbers selected; (c) at least 6 of the numbers selected?

A 3-person basketball team consists of a guard, a forward, and a center. (a) If a person is chosen at random from each of three different such teams, what is the probability of selecting a complete team? (b) What is the probability that all 3 players selected play the same position?

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired; (b) Rebecca and Elise will be chosen to represent their schools but will not play each other; (c) exactly one of Rebecca and Elise will be chosen to represent her school?

Use induction to generalize Bonferroni's inequality to \(n\) events. Namely, show that $$ P\left(E_{1} E_{2} \cdots E_{n}\right) \geq P\left(E_{1}\right)+\cdots+P\left(E_{n}\right)-(n-1) $$

If 8 castles (that is, rooks) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Probability and 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.