/*! 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 28 An urn contains 5 red, 6 blue, a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 28

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement.

Short Answer

Expert verified
In summary, for the given problem, we have: 1. Without Replacement: (a) Probability of all 3 balls being of the same color: \( \frac{86}{969} \) (b) Probability of all 3 balls being of different colors: \( \frac{240}{969} \) 2. With Replacement: (a) Probability of all 3 balls being of the same color: (5/19)^3 + (6/19)^3 + (8/19)^3 (b) Probability of all 3 balls being of different colors: \( 6 * (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \)

Step by step solution

01

Calculate the total number of ways to select 3 balls from the urn

There are a total of 5 (red) + 6 (blue) + 8 (green) = 19 balls in the urn. We can choose 3 balls in C(19,3) ways, where C(n,r) denotes the number of possible combinations of choosing 'r' items from a set of 'n' items. Using the combination formula, we have: C(19,3) = \( \frac{19!}{3!(19-3)!} = \frac{19!}{3!16!} = 969\).
02

Calculate the probability of (a) all 3 balls being of the same color

This event can occur in 3 ways: all red, all blue, or all green. - All red: We can select 3 red balls in C(5,3) ways, i.e. \( \frac{5!}{3!2!} = 10\). - All blue: We can select 3 blue balls in C(6,3) ways, i.e. \( \frac{6!}{3!3!} = 20\). - All green: We can select 3 green balls in C(8,3) ways, i.e. \( \frac{8!}{3!5!} = 56\). The probability of this event is the sum of these possibilities divided by the total number of ways to select 3 balls, i.e., \( P(A) = \frac{10+20+56}{969} = \frac{86}{969}\).
03

Calculate the probability of (b) all 3 balls being of different colors

For this event to occur, we must choose one ball of each color. The number of ways to choose one ball from each color is: 5 (for red) * 6 (for blue) * 8 (for green) = 240. The probability of this event is the number of ways to choose one ball of each color divided by the total number of ways to select 3 balls, i.e., \( P(B) = \frac{240}{969}\). #Case 2: With Replacement#
04

Calculate the probability of (a) all 3 balls being of the same color, with replacement

For each ball, we have a probability of choosing: - Red: \( \frac{5}{19} \) - Blue: \( \frac{6}{19} \) - Green: \( \frac{8}{19} \) Since the balls are replaced after noting their color, the probability of choosing all 3 balls of the same color is the sum of probabilities for all red, all blue, or all green: (5/19)^3 + (6/19)^3 + (8/19)^3.
05

Calculate the probability of (b) all 3 balls being of different colors, with replacement

We can have 3! = 6 arrangements for the 3 balls of different colors. For each arrangement, the probability of that arrangement occurring is the product of the probabilities for each color, i.e., \( (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \). Since there are 6 such arrangements, the overall probability for this case is 6 times the above expression: \( 6 * (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \).

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

The game of craps is played as follows: A player rolls two dice. If the sum of the dice is either a \(3,\) or \(12,\) the player loses; if the sum is either a 7 or an \(11,\) the player wins. If the outcome is anything else, the player continues to roll the dice until she rolls either the initial outcome or a \(7 .\) If the 7 comes first, the player loses, whereas if the initial outcome reoccurs before the 7 appears, the player wins. Compute the probability of a player winning at craps. Hint: Let \(E_{i}\) denote the event that the initial outcome is \(i\) and the player wins. The desired probability is \(\sum_{i=2}^{12} P\left(E_{i}\right) .\) To compute \(P\left(E_{i}\right),\) define the events \(E_{i, n}\) to be the event that the initial sum is \(i\) and the player wins on the \(n\) th roll. Argue that \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\) 2 events \(E_{i, n}\) to be the event that the initial sum is \(i\) and the player wins on the \(n\) th roll. Argue that \(P\left(E_{i}\right)=\sum_{n=1}^{\infty} P\left(E_{i, n}\right)\).

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace? (b) a ring and a necklace?

A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be (a) no complete pair? (b) exactly 1 complete pair?

If two dice are rolled, what is the probability that the sum of the upturned faces equals \(i ?\) Find it for \(i=2,3, \ldots, 11,12\).

There are 5 hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel? What assumptions are you making?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.