/*! 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 20 A customer at Cavallaro's Fruit ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A customer at Cavallaro's Fruit Stand picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotten. What is the probability that the sample

Short Answer

Expert verified
Probability (At least one rotten orange) = 1 - \( \frac{C(56, 3)}{C(60, 3)} \) = 1 - \( \frac{\frac{56!}{3!(56-3)!}}{\frac{60!}{3!(60-3)!}} \) Simplify the equation to get: Probability (At least one rotten orange) = 1 - \( \frac{56! \cdot 3! \cdot (60-3)!}{60! \cdot 3! \cdot (56-3)!} \) = 1 - \( \frac{56!}{60!} \) Probability (At least one rotten orange) ≈ 0.288

Step by step solution

01

Determine the total number of combinations

First, we must find the total number of ways to draw 3 oranges out of 60. This is a combination problem, which can be calculated using the formula for combinations, denoted as "C(n, r)", where n is the total number of items and r is the number of items to choose. In this case, n = 60 and r = 3. Calculate C(60, 3) = \( \frac{60!}{3!(60-3)!} \)
02

Determine the number of combinations with no rotten oranges

Next, we need to find the total number of ways to choose 3 oranges out of the 56 good ones, since there are 60 - 4 = 56 good oranges. Calculate C(56, 3) = \( \frac{56!}{3!(56-3)!} \)
03

Calculate the probability of choosing no rotten oranges

Now that we have the total number of combinations of 3 oranges out of 60 and the total number of combinations with no rotten oranges, we can calculate the probability of drawing 3 good oranges. Probability = \( \frac{\text{Number of good combinations}}{\text{Total number of combinations}} \) Probability (No rotten oranges) = \( \frac{C(56, 3)}{C(60, 3)} \)
04

Calculate the complement probability

The probability of drawing at least one rotten orange is the complement of drawing no rotten oranges. We can find this probability by subtracting the probability of drawing no rotten oranges from 1. Probability (At least one rotten orange) = 1 - Probability (No rotten oranges)
05

Final Answer

Plug in the calculated values for C(56, 3) and C(60, 3) into the respective equations from Steps 3 and 4 and simplify to find the final probability of choosing a sample that contains at least one rotten orange.

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.

Combinations
Combinations are all about selecting a group from a larger set, without considering the order in which they were picked. In the world of mathematics, this is a handy way to figure out how many ways we can make a selection. When you're given a total number and have to choose a few from them, think of it as making a combination.
The formula we use to calculate combinations is written as \( C(n, r) \), where \( n \) is the total number of items, and \( r \) is the number you want to choose. This is represented mathematically as:
  • \( C(n, r) = \frac{n!}{r!(n-r)!} \)
Here, "!" represents factorial, which is the product of a whole number and all the numbers below it. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).
Using combinations helps in scenarios like picking a committee, dealing cards, or in our exercise, choosing fruit from a crate. It's important to note that since order doesn't matter, choosing an apple and then an orange is identical to choosing an orange and then an apple.
Discrete Mathematics
Discrete Mathematics is a branch of math that deals with objects that can take on only distinct, separate values. It's like focusing on the dots rather than the lines that connect them. This includes things like integers, graphs, and statements in logic.
One of the key areas in discrete mathematics is combinatorics, which is all about counting the ways certain situations can unfold. For instance, determining how many different ways you can arrange flowers in a vase, or the way you can pick teams for a game, is part of combinatorial calculations.
Discrete mathematics helps in various fields, like computer science, where algorithms need exact steps, or operations research, where resources are allocated optimally. It's highly applicable in coding, encryption, and networks. In the exercise we're dealing with, picking out oranges applies combinatorial principles, a fundamental area within discrete mathematics.
Complement Rule
The Complement Rule is a nifty trick in probability that helps find chances of an event by looking at the possibility of it not happening. This rule is like asking, "What's the probability of not losing the game?" instead of asking, "What's the probability of winning?"
Mathematically, it’s expressed as:
  • Probability of Event (A) = 1 - Probability of Event (A not happening)
It's very useful when calculating the probability directly is tough, but calculating the opposite case is easier as seen in the exercise.
In our fruit stand example, instead of figuring out how likely it is to get at least one rotten orange, we see how likely it is to get only good ones. Then, using the complement rule, subtract this probability from 1, giving the probability of getting at least one rotten orange. This rule simplifies tasks, especially in complicated problems where determining every possible outcome is challenging.

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 admissions office of a private university released the following admission data for the preceding academic year: From a pool of 3900 male applicants, \(40 \%\) were accepted by the university and \(40 \%\) of these subsequently enrolled. Additionally, from a pool of 3600 female applicants, \(45 \%\) were accepted by the university and \(40 \%\) of these subsequently enrolled. What is the probability that a. A male applicant will be accepted by and subsequently will enroll in the university? b. A student who applies for admissions will he accepted by the university? c. A student who applies for admission will be accepted by the university and subsequently will enroll?

Suppose the probability that Bill can solve a problem is \(p_{1}\) and the probability that Mike can solve it is \(p_{2}\). Show that the probability that Bill and Mike working independently can solve the problem is \(p_{1}+p_{2}-p_{1} p_{2}\).

A time study was conducted by the production manager of Universal Instruments to determine how much time it took an assembly worker to complete a certain task during the assembly of its Galaxy home computers. Results of the study indicated that \(20 \%\) of the workers were able to complete the task in less than \(3 \mathrm{~min}\), \(60 \%\) of the workers were able to complete the task in \(4 \mathrm{~min}\) or less, and \(10 \%\) of the workers required more than \(5 \mathrm{~min}\) to complete the task. If an assembly-line worker is selected at random from this group, what is the probability that a. He or she will be able to complete the task in 5 min or less? b. He or she will not be able to complete the task within 4 min? c. The time taken for the worker to complete the task will be between 3 and 4 min (inclusive)?

Quaurr CoNrroL. An automobile manufacturer obtains the microprocessors used to regulate fuel consumption in its automobiles from three microelectronic firms: \(\mathrm{A}, \mathrm{B}\), and C. The quality-control department of the company has determined that \(1 \%\) of the microprocessors produced by firm \(A\) are defective, \(2 \%\) of those produced by firm \(B\) are defective, and \(1.5 \%\) of those produced by firm \(\mathrm{C}\) are defective. Firms \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) supply \(45 \%, 25 \%\), and \(30 \%\), respectively, of the microprocessors used by the company. What is the probability that a randomly selected automobile manufactured by the company will have a defective microprocessor?

Fifty raffle tickets are numbered 1 through 50 , and one of them is drawn at random. What is the probability that the number is a multiple of 5 or \(7 ?\) Consider the following "solution": Since 10 tickets bear numbers that are multiples of 5 and since 7 tickets bear numbers that are multiples of 7 , we conclude that the required probability is $$ \frac{10}{50}+\frac{7}{50}=\frac{17}{50} $$ What is wrong with this argument? What is the correct answer?

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.