/*! 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 26 Yahtzee (5.3,6.3) In the game of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Yahtzee (5.3,6.3) In the game of Yahtzee, 5 sixsided dice are rolled simultaneously. To get a Yahtzee, the player must get the same number on all 5 dice. (a) Luis says that the probability of getting a Yahtzee in one roll of the dice is \(\left(\frac{1}{6}\right)^{5} .\) Explain why Luis is wrong. (b) Nassir decides to keep rolling all 5 dice until he gets a Yahtzee. He is surprised when he still hasn't gotten a Yahtzee after 25 rolls. Should he be? Calculate an appropriate probability to support your answer.

Short Answer

Expert verified
(a) Luis is wrong; the probability should be \(\frac{6}{1296}\). (b) Nassir shouldn't be surprised; the probability of not getting a Yahtzee in 25 rolls is 98.1%.

Step by step solution

01

Understanding Probability of Yahtzee

To obtain a Yahtzee, all 5 dice must show the same number. Each die has 6 possibilities, so there are 6 successful outcomes (all dice showing 1, all showing 2, etc.). The total number of outcomes when rolling 5 dice is \(6^5\). Thus, the probability of getting a Yahtzee is \(\frac{6}{6^5} = \frac{1}{1296}\).
02

Analyzing Luis' Calculation

Luis calculated the probability as \(\left(\frac{1}{6}\right)^5\). This would be the probability of obtaining all dice showing one specific number (say all 1s) but doesn't account for all six possible outcomes (all 1s, all 2s, etc.). He should multiply \(\left(\frac{1}{6}\right)^5\) by 6, making his calculation incorrect.
03

Calculating the Probability of Not Getting a Yahtzee in 25 Rolls

The probability of not getting a Yahtzee in a single roll is \(1 - \frac{1}{1296}\). Raise this probability to the 25th power to account for 25 independent rolls: \(\left(1 - \frac{1}{1296}\right)^{25}\).
04

Calculating the Probability Using a Calculator

Calculating \(\left(1 - \frac{1}{1296}\right)^{25}\) yields approximately 0.9810. This means there is roughly a 98.1% chance of not rolling a Yahtzee in 25 tries.
05

Conclusion Regarding Nassir's Surprise

The probability derived indicates a 1.9% chance of obtaining a Yahtzee in 25 rolls. This is very low, so Nassir should not be surprised that he hasn't rolled a Yahtzee yet.

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.

Yahtzee Probability
Yahtzee is a game that offers an exciting way to explore probability, especially when computing the likelihood of achieving a rare outcome known as a "Yahtzee." In the game, a Yahtzee occurs when all five six-sided dice show the identical number. The probability of obtaining a Yahtzee in a single roll of the dice is calculated by considering both the specific and overall outcomes. Each die has 6 faces, leading to 6 possible favorable outcomes (like all dice showing one number).

The total number of possible outcomes for rolling 5 dice at once is calculated as \(6^5\) because each die can land on any of the 6 sides. Thus, the probability of rolling a Yahtzee is given by the ratio of favorable outcomes to the total outcomes: \(\frac{6}{6^5} = \frac{1}{1296}\). This illustrates that getting a Yahtzee is indeed a rare event, contributing to the thrill and challenge of the game.
Calculating Outcomes
Calculating the probability of obtaining specific outcomes in probability exercises, such as Yahtzee, involves recognizing all potential possibilities. The misstep Luis made in the exercise was calculating the probability for a single specific outcome, rather than considering all equivalent outcomes.
  • The probability of a specific set of results, such as all dice showing "1", is \(\left(\frac{1}{6}\right)^5\).
  • However, since any number (1 through 6) could fit this scenario, we need to account for all possible successes, thus multiplying by 6 gives us \(6 \times \left(\frac{1}{6}\right)^5\).
Recognizing the full set of possible favorable outcomes is crucial to ensure the accuracy of probability calculations. It emphasizes the necessity of thinking through all equivalent scenarios when calculating the likelihood of an event in games and other contexts.
Independent Rolls
The principle of independent events is essential in understanding probability, especially in scenarios involving multiple trials, such as repeated dice rolls in Yahtzee. Independent rolls mean that the outcome of one roll does not affect the outcome of another roll. This is significant when calculating probabilities over several trials.

In the context of Nassir's 25 dice rolls, the primary consideration is that each roll is an independent event. The probability of not achieving a Yahtzee in a single roll is \(1 - \frac{1}{1296}\). To determine the probability of failing to get a Yahtzee over 25 rolls, this probability is multiplied for each independent roll: \(\left(1 - \frac{1}{1296}\right)^{25}\).

This concept shows how probabilities of sequences of events are calculated, stressing the independent nature of each event in the sequence and allowing for precise probabilistic predictions.
Probability Calculation Errors
Probability calculations can sometimes contain errors, as seen in Luis' initial mistake, often resulting from not accounting for all possible favorable outcomes in a situation. Here are some tips to avoid common probability calculation errors:
  • Ensure that every possible favorable outcome has been considered. In cases like Yahtzee, explore all configurations that result in a success.
  • Understand the rules of the game or scenario to properly identify all potential outcomes.
  • Practice breaking down complex scenarios into simpler parts to help uncover any overlooked probabilities.
Avoiding these errors involves careful thought, practice, and an understanding of the basic principles of probability. By doing so, one can calculate probabilities accurately and avoid making erroneous assumptions or simplifications.

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

How are schools doing? The nonprofit group Public Agenda conducted telephone interviews with three randomly selected groups of parents of high school children. There were 202 black parents, 202 Hispanic parents, and 201 white parents. One question asked was "Are the high schools in your state doing an excellent, good, fair, or poor job, or don't you know enough to say?" Here are the survey results: \({ }^{14}\) $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Black } \\ \text { parents } \end{array} & \begin{array}{c} \text { Hispanic } \\ \text { parents } \end{array} & \begin{array}{c} \text { White } \\ \text { parents } \end{array} \\ \text { Excellent } & 12 & 34 & 22 \\ \text { Good } & 69 & 55 & 81 \\ \text { Fair } & 75 & 61 & 60 \\ \text { Poor } & 24 & 24 & 24 \\ \text { Don't know } & 22 & 28 & 14 \\ \text { Total } & 202 & 202 & 201 \\ \hline \end{array} $$ (a) Calculate the conditional distribution (in proportions) of responses for each group of parents. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of responses for the three groups of parents.

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ (a) A Type I error is possible. (b) A Type II error is possible. (c) Both a Type I and a Type II error are possible. (d) There is no chance of making a Type I or Type II error because the \(P\) -value is approximately \(0 .\) (e) There is no chance of making a Type I or Type II error because the calculations are correct. For these data, \(\chi^{2}=69.8\) with a \(P\) -value of approximately \(0 .\) Assuming that the researchers used a significance level of \(0.05,\) which of the following is true?

How to quit smoking It's hard for smokers to quit. Perhaps prescribing a drug to fight depression will work as well as the usual nicotine patch. Perhaps combining the patch and the drug will work better than either treatment alone. Here are data from a randomized, double-blind trial that compared four treatments. \({ }^{19} \mathrm{~A}\) "success" means that the subject did not smoke for a year following the beginning of the study. $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Subjects } & \text { Successes } \\ 1 & \text { Nicotine patch } & 244 & 40 \\ 2 & \text { Drug } & 244 & 74 \\ 3 & \text { Patch plus drug } & 245 & 87 \\ 4 & \text { Placebo } & 160 & 25 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the success rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=\) \(p_{2}=p_{3}=p_{4}\) says about subjects' smoking habits. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{3}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ A forensic accountant who is familiar with Benford's law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data. $$ \begin{array}{lcrrrrrrrr} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Count: } & 61 & 50 & 43 & 34 & 25 & 16 & 7 & 8 & 6 \\ \hline \end{array} $$ (a) Are these data inconsistent with Benford's law? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

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.