/*! 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 51 In the 18th century, the Chevali... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the 18th century, the Chevalier de Mere asked Blaise Pascal to compare the probabilities of two events. Below, you will compute the probability of the two events that, prior to contrary gambling experience, were thought by de Mere to be equally likely. a. What is the probability of obtaining at least one 6 in four rolls of a fair die? Answer \(t\) b. If a pair of fair dice is tossed 24 times, what is the probability of at least one double six?

Short Answer

Expert verified
Event (a) probability: \( \frac{671}{1296} \); Event (b) probability: \( 0.4914 \).

Step by step solution

01

Understanding the Problem

We need to compare probabilities of two events involving dice rolls: (a) obtaining at least one 6 in four rolls of a single die, and (b) getting at least one double six in 24 rolls of a pair of dice.
02

Calculate Probability (Event A)

For event (a), we need the probability of getting at least one '6' in four rolls of a single die. First, calculate the probability of not getting a '6' in a single roll, which is \( \frac{5}{6} \). Then, the probability of not rolling a '6' in four rolls is \( \left( \frac{5}{6} \right)^4 \). The probability of getting at least one '6' is the complement: \( 1 - \left( \frac{5}{6} \right)^4 \).
03

Calculate Result for Event A

Now compute \( \left( \frac{5}{6} \right)^4 \) which equals \( \frac{625}{1296} \). Therefore, the probability of getting at least one '6' is \( 1 - \frac{625}{1296} = \frac{671}{1296} \).
04

Calculate Probability (Event B)

For event (b), we need the probability of getting at least one double six in 24 rolls of two dice. The probability for a double six in a single roll of two dice is \( \frac{1}{36} \). Therefore, the probability of not rolling a double six is \( \frac{35}{36} \). In 24 rolls, the probability of not getting a double six is \( \left( \frac{35}{36} \right)^{24} \). The probability of getting at least one double six is the complement: \( 1 - \left( \frac{35}{36} \right)^{24} \).
05

Calculate Result for Event B

Compute \( \left( \frac{35}{36} \right)^{24} \). This value is approximately 0.5086; hence, the probability of getting at least one double six is \( 1 - 0.5086 = 0.4914 \).

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
Understanding dice probability is essential when dealing with problems involving dice rolls. A standard die is a cube with six faces, each showing a different number from 1 to 6. In probability theory, each face is equally likely to appear on each roll. This leads to a fundamental rule: the probability of any specific outcome is \[ \frac{1}{6} \].

When computing probabilities that involve rolling dice multiple times, it’s crucial to consider events such as 'at least one occurrence.' For instance, the probability of rolling at least one six in four rolls involves computing the chance of not getting a six in all four rolls first. Then, you apply complementary probability to find out 'at least one' cases. This exemplifies how probabilities build up when multiple trials are involved.
Combinatorics
Combinatorics is a field of mathematics focused on counting and arranging sets. In probability theory, combinatorics helps determine the number of possible outcomes and favorable outcomes in an event. It often involves calculating combinations and permutations.

In our dice problem, combinatorics provides the method to understand permutations of dice rolls. If we were considering multiple dice, such as two dice being rolled at once, combinatorics helps us determine probabilities of different results – like calculating the chance of a double six out of many trials. The clear understanding of permutation and combination allows computing probabilities in complex scenarios accurately.
Pascal's Problem
Blaise Pascal, a prominent mathematician, dealt with many probability challenges in his time, often around games of chance. Pascal's problem as stated, compares two dice-based events to determine which is more likely. This is a classic probability question: predicting outcomes with dice rolled multiple times.

His approach demonstrates the need for understanding both simple dice probabilities and more complex calculations involving multiple trials and outcomes. Pascal’s insight was to compare complementary probabilities to solve real-world problems of gambling and chance, providing a foundation for modern probability theory and statistical analysis.
Complementary Probability
The concept of complementary probability plays a pivotal role in solving problems involving "at least one" scenarios. It’s based on the idea that the probabilities of all possible outcomes of an experiment must add up to one.

In practical terms, if you want to find the probability of an event happening at least once, you can calculate the probability of the event not happening and subtract this from one. For example, in dice rolls, finding out 'at least one six' within several rolls is determined by calculating the chance of rolling no sixes at all in those rolls, and then using the formula \[ 1 - P(\text{not rolling a six}) \].

This methodology reduces complex probability questions to more manageable calculations, leveraging the fact that probabilities of complementary events mirror each other.

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 southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three are randomly selected to fill the open positions. Let \(Y\) be the number of internship trained candidates who are hired. a. Does \(Y\) have a binomial or hypergeometric distribution? Why? b. Find the probability that two or more internship trained candidates are hired. c. What are the mean and standard deviation of \(Y\) ?

A particular sale involves four items randomly selected from a large lot that is known to contain 10\% defectives. Let \(Y\) denote the number of defectives among the four sold. The purchaser of the items will return the defectives for repair, and the repair cost is given by \(C=3 Y^{2}+Y+2 .\) Find the expected repair cost. [Hint: The result of Theorem 3.6 implies that, for any random variable \(Y\), \(\left.E\left(Y^{2}\right)=\sigma^{2}+\mu^{2} \cdot\right].\)

It is known that \(5 \%\) of the members of a population have disease \(A,\) which can be discovered by a blood test. Suppose that \(N\) (a large number) people are to be tested. This can be done in two ways: 1\. Each person is tested separately, or 2\. the blood samples of \(k\) people are pooled together and analyzed. (Assume that \(N=n k\), with \(n\) an integer.) If the test is negative, all of them are healthy (that is, just this one test is needed). If the test is positive, each of the \(k\) persons must be tested separately (that is, a total of \(k+1\) tests are needed). a. For fixed \(k,\) what is the expected number of tests needed in option \(2 ?\) b. Find the \(k\) that will minimize the expected number of tests in option 2 . c. If \(k\) is selected as in part (b), on the average how many tests does option 2 save in comparison with option 1?

a. We observe a sequence of independent identical trials with two possible outcomes on each trial, \(S\) and \(F,\) and with \(P(S)=p .\) The number of the trial on which we observe the fifth success, \(Y\), has a negative binomial distribution with parameters \(r=5\) and \(p\). Suppose that we observe the fifth success on the eleventh trial. Find the value of \(p\) that maximizes \(P(Y=11\) ). b. Generalize the result from part (a) to find the value of \(p\) that maximizes \(P\left(Y=y_{0}\right)\) when \(Y\) has a negative binomial distribution with parameters \(r\) (known) and \(p\)

A city commissioner claims that \(80 \%\) of the people living in the city favor garbage collection by contract to a private company over collection by city employees. To test the commissioner's claim, 25 city residents are randomly selected, yielding 22 who prefer contracting to a private company. a. If the commissioner's claim is correct, what is the probability that the sample would contain at least 22 who prefer contracting to a private company? b. If the commissioner's claim is correct, what is the probability that exactly 22 would prefer contracting to a private company? c. Based on observing 22 in a sample of size 25 who prefer contracting to a private company, what do you conclude about the commissioner's claim that \(80 \%\) of city residents prefer contracting to a private company?

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.