/*! 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 die is rolled twice. What is t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A die is rolled twice. What is the probability that the sum of the faces is greater than 7 , given that (a) the first outcome was a \(4 ?\) (b) the first outcome was greater than \(3 ?\) (c) the first outcome was a \(1 ?\) (d) the first outcome was less than \(5 ?\)

Short Answer

Expert verified
(a) \( \frac{1}{2} \) (b) \( \frac{2}{3} \) (c) 0 (d) \( \frac{1}{4} \)

Step by step solution

01

Understand the Problem

We need to find probabilities related to the sum of two dice rolls under different conditions. Specifically, we want the probability that the sum is greater than 7 given certain conditions about the first die roll.
02

Calculate Possible Outcomes for a Die Roll

When one six-sided die is rolled, there are 6 possible outcomes: 1, 2, 3, 4, 5, and 6. Thus, if a die is rolled twice, there are 36 possible outcomes in total for these two rolls.
03

(a): Probability, Given First Outcome is 4

Given the first die rolled a 4, the second die rolls could be 1 through 6. Calculate the cases where the sum is greater than 7: (4, 4), (4, 5), (4, 6). There are 3 successful outcomes out of 6 possibilities.Hence, the probability is \[P = \frac{3}{6} = \frac{1}{2}.\]
04

(b): Probability, Given First Outcome is Greater than 3

The possible outcomes for the first die are 4, 5, and 6. For:- First die 4: (4, 4), (4, 5), (4, 6) are successful.- First die 5: (5, 3), (5, 4), (5, 5), (5, 6) are successful.- First die 6: (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) are successful.That makes 3 + 4 + 5 = 12 successful outcomes out of 18 possibilities.So, the probability is \[P = \frac{12}{18} = \frac{2}{3}.\]
05

(c): Probability, Given First Outcome is 1

The possible outcomes for the second die are 1 through 6. None of these results in a sum greater than 7. Therefore, there are 0 successful outcomes out of 6 possibilities.Thus, the probability is zero: \[P = \frac{0}{6} = 0.\]
06

(d): Probability, Given First Outcome is Less than 5

The possible outcomes for the first die are 1, 2, 3, and 4. For:- First die 1: 0 successful outcomes.- First die 2: (2, 6) is successful.- First die 3: (3, 5), (3, 6) are successful.- First die 4: (4, 4), (4, 5), (4, 6) are successful.That makes 0 + 1 + 2 + 3 = 6 successful outcomes out of 24 possibilities.So, the probability is \[P = \frac{6}{24} = \frac{1}{4}.\]

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 91Ó°ÊÓ!

Key Concepts

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

Dice Probability
When we talk about dice probability, we mean the chance of a specific event happening when we roll one or more dice. A standard die is a six-sided cube, with each side having an equal chance of landing face up, making the probability of each face \( \frac{1}{6} \).
When rolling two dice, we often consider the combined outcomes. Since a single die has 6 faces, rolling two dice results in \( 6 \times 6 = 36 \) possible outcomes.
This foundational understanding is crucial for solving problems involving dice, as it forms the base for building more complex probability scenarios.
Sum of Dice
The sum of dice is a vital concept in calculating probabilities involving multiple dice. When rolling two dice, sums can range from 2 (1+1) to 12 (6+6).
Each sum has several possibilities, and some sums are more likely than others, such as 7, because they can be achieved in multiple ways (e.g., 1+6, 2+5, 3+4, etc.).
Understanding the distribution of these sums is essential. It impacts how we determine conditional probabilities, like finding the chance of a sum being greater than 7, depending on prior constraints about the dice outcomes.
Probability Calculation
Probability calculation is about determining the likelihood of a particular outcome or set of outcomes. It is often expressed as a fraction, with the number of successful outcomes over the total number of possible outcomes.
For example, if you want to find the probability that the sum of two dice is greater than 7, you start by identifying all the possible successful outcomes. Then you divide that number by the total number of outcomes, which is always 36 for two dice.
In conditional probability, like in the given exercise, we further limit our universe by some condition (e.g., the first die is a 4). This changes the potential outcomes we consider, tailoring the calculation to this restricted set.
Conditional Outcomes
Conditional outcomes allow us to find probabilities when additional information is provided. In our exercise, given conditions such as "the first roll is 4" change how we approach our probability problem.
With the first roll known, the number of potential outcomes adjusts from 36 to 6 because only the second die's outcome is now uncertain. This makes our calculations more precise
The step-by-step solutions highlight this by recalculating the probability for specific conditions, showing how focused information can simplify complex probability scenarios. This helps us see that probability isn't static but shifts with each piece of new information.

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

George Wolford has suggested the following variation on the Linda problem (see Exercise 1.2.25). The registrar is carrying John and Mary's registration cards and drops them in a puddle. When he pickes them up he cannot read the names but on the first card he picked up he can make out Mathematics 23 and Government \(35,\) and on the second card he can make out only Mathematics \(23 .\) He asks you if you can help him decide which card belongs to Mary. You know that Mary likes government but does not like mathematics. You know nothing about John and assume that he is just a typical Dartmouth student. From this you estimate: \(P(\) Mary takes Government 35\()=.5\) \(P(\) Mary takes Mathematics 23\()=.1\) \(P(\) John takes Government 35\()=.3\) \(P(\) John takes Mathematics 23\()=.2\) Assume that their choices for courses are independent events. Show that the card with Mathematics 23 and Government 35 showing is more likely to be Mary's than John's. The conjunction fallacy referred to in the Linda problem would be to assume that the event "Mary takes Mathematics 23 and Government \(35 "\) is more likely than the event "Mary takes Mathematics \(23 . "\) Why are we not making this fallacy here?

What is the probability that a family of two children has (a) two boys given that it has at least one boy? (b) two boys given that the first child is a boy?

Let \(\Omega=\\{a, b, c, d, e, f\\} .\) Assume that \(m(a)=m(b)=1 / 8\) and \(m(c)=\) \(m(d)=m(e)=m(f)=3 / 16 .\) Let \(A, B,\) and \(C\) be the events \(A=\\{d, e, a\\}\) \(B=\\{c, e, a\\}, C=\\{c, d, a\\} .\) Show that \(P(A \cap B \cap C)=P(A) P(B) P(C)\) but no two of these events are independent.

Suppose that \(X\) and \(Y\) are continuous random variables with density functions \(f_{X}(x)\) and \(f_{Y}(y)\), respectively. Let \(f(x, y)\) denote the joint density function of \((X, Y)\). Show that $$ \int_{-\infty}^{\infty} f(x, y) d y=f_{X}(x) $$ and $$ \int_{-\infty}^{\infty} f(x, y) d x=f_{Y}(y) $$

Suppose you toss a dart at a circular target of radius 10 inches. Given that the dart lands in the upper half of the target, find the probability that (a) it lands in the right half of the target. (b) its distance from the center is less than 5 inches. (c) its distance from the center is greater than 5 inches. (d) it lands within 5 inches of the point (0,5) .

See all solutions

Recommended explanations on Math Textbooks

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.