/*! 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 17 A pair of fair dice is rolled. W... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 17

A pair of fair dice is rolled. What is the probability that the sum of the numbers falling uppermost is less than 9, given that at least one of the numbers is a 6 ?

Short Answer

Expert verified
The probability that the sum of the numbers falling uppermost is less than 9, given that at least one of the numbers is a 6, is \(\frac{7}{11}\).

Step by step solution

01

Identify the possible outcomes with at least one 6

Since there are 6 possible faces for each die, the total number of outcomes when rolling a pair of dice is 6 * 6 = 36. However, since we need to consider only those outcomes where at least one of the dice shows a 6, we will eliminate the cases where neither die has a 6. There are 5 possible faces (1 through 5) for each die in this case, so the total number of outcomes without a 6 is 5 * 5 = 25. Therefore, the number of possible outcomes with at least one 6 is 36 - 25 = 11.
02

Identify the favorable outcomes

Now we need to find the number of favorable outcomes, i.e., those outcomes with a sum less than 9 and at least one 6. There are two possibilities here: either one die has a 6 and the other has a number from 1 to 3, or both dice have a number 6. The former case has 3 possible favorable outcomes for each die (since there are 3 numbers from 1 to 3). Since there are two dice, this gives us a total of 3 + 3 = 6 favorable outcomes for this case. In the latter case, there is only 1 favorable outcome (both dice have a 6). So, the total number of favorable outcomes is 6 + 1 = 7.
03

Calculate the probability

Now that we have determined the number of favorable outcomes and the total number of possible outcomes with at least one 6, we can calculate the probability of obtaining a sum less than 9 given that at least one of the dice shows a 6. This probability is equal to the quotient of the number of favorable outcomes divided by the total number of possible outcomes with at least one 6. In this case, the probability is given by 7 (favorable outcomes) divided by 11 (possible outcomes with at least one 6). Therefore, the probability is \(\frac{7}{11}\). So, the probability that the sum of the numbers falling uppermost is less than 9, given that at least one of the numbers is a 6, is \(\frac{7}{11}\).

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!

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

Applicants for temporary office work at Carter Temporary Help Agency who have successfully completed a typing test are then placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temporary help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience it is known that \(80 \%\) of the employees placed by Nancy are rated as satisfactory, and \(70 \%\) of those placed by Darla are rated as satisfactory. Darla places \(55 \%\) of the temporary office help at the agency and Nancy the remaining \(45 \%\). If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla?

The estimated probability that a brand-A, a brand-B, and a brand-C plasma TV will last at least \(30,000 \mathrm{hr}\) is \(.90, .85\), and \(.80\), respectively. Of the 4500 plasma TVs that Ace TV sold in a certain year, 1000 were brand A, 1500 were brand \(\mathrm{B}\), and 2000 were brand \(\mathrm{C}\). If a plasma TV set sold by Ace TV that year is selected at random and is still working after \(30,000 \mathrm{hr}\) of use a. What is the probability that it was a brand-A TV? b. What is the probability that it was not a brand-A TV?

A study of the faculty at U.S. medical schools in 2006 revealed that \(32 \%\) of the faculty were women and \(68 \%\) were men. Of the female faculty, \(31 \%\) were full/ associate professors, \(47 \%\) were assistant professors, and \(22 \%\) were instructors. Of the male faculty, \(51 \%\) were full/associate professors, \(37 \%\) were assistant professors, and \(12 \%\) were instructors. If a faculty member at a U.S. medical school selected at random holds the rank of full/associate professor, what is the probability that she is female?

Three cards are drawn without replacement from a wellshuffled deck of 52 playing cards. What is the probability that the third card drawn is a diamond?

Let \(E\) and \(F\) be events such that \(F \subset E\). Find \(P(E \mid F)\) and interpret your result.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

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.