/*! 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 106 Spinning for apples (6.3 or 7.3)... [FREE SOLUTION] | 91影视

91影视

Spinning for apples (6.3 or 7.3) In the 鈥淎sk Marilyn鈥 column of Parade magazine, a reader posed this question: 鈥淪ay that a slot machine has five wheels, and each wheel has five symbols: an apple, a grape, a peach, a pear, and a plum. I pull the lever five times. What are the chances that I鈥檒l get at least one apple?鈥 Suppose that the wheels spin independently and that the five symbols are equally likely to appear on each wheel in a given spin. (a) Find the probability that the slot player gets at least one apple in one pull of the lever. Show your method clearly. (b) Now answer the reader鈥檚 question. Show your method clearly.

Short Answer

Expert verified
(a) Probability of getting at least one apple in one spin is \( \frac{2101}{3125} \); (b) In five spins, it's approximately 0.965.

Step by step solution

01

Define the Probability of Not Getting an Apple

First, determine the probability that a single symbol on a single wheel is not an apple. Since there are five equally likely symbols and only one of them is an apple, there are four symbols that are not apples out of five. This probability can be calculated as \( P( ext{Not Apple}) = \frac{4}{5} \).
02

Probability of No Apples in One Spin

Since there are five wheels, and each has an independent chance of showing a non-apple symbol, calculate the probability of getting no apples across all five wheels in a single spin. This is \( \left( \frac{4}{5} \right)^5 = \frac{1024}{3125} \).
03

Complement Rule to Find At Least One Apple

To find the probability of getting at least one apple, use the complement rule. Subtract the probability of not getting an apple from 1: \( P( ext{at least one apple}) = 1 - \left( \frac{1024}{3125} \right) = \frac{2101}{3125} \).
04

Probability for Five Spins

To determine the probability of getting at least one apple in five separate spins, first find the probability of not getting any apples in one spin and then raise it to the power of 5. We have \( P( ext{No apples in five spins}) = \left( \frac{1024}{3125} \right)^5 \). After calculation, \( P( ext{No apples in five spins}) \approx 0.035 \).
05

Complement Rule for Five Spins

Using the complement rule again, find the probability of getting at least one apple in one of the five spins. Thus, \( P( ext{at least one apple in five spins}) = 1 - (0.035) = 0.965 \).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complement Rule
The Complement Rule is a fundamental principle in probability. It helps us find the probability of an event happening by using the probability of the event not happening. When you're tasked with finding the probability of getting at least one apple in our slot machine problem, this rule becomes handy.

To grasp the concept, let's consider a simple equation:
  • If the total is 1 (representing certainty), then the probability of an event plus the probability of its complement (the opposite event) must equal 1.
This means:
  • 1 = P(event happens) + P(event doesn't happen)
For example, if you know the probability of not getting any apples across all five wheels is approximately 0.328 (or about 32.8%), you can calculate the probability of getting at least one apple as 1 - 0.328, which equals 0.672 (67.2%).

By using the Complement Rule, even complex probability questions become simpler. This method reduces the chance of overlooking possible outcomes.
Independent Events
Understanding Independent Events is crucial when dealing with probability, especially in problems involving multiple processes like the slot machine. Essentially, events are independent if the occurrence of one does not affect the probability of the other.

In our problem, when you pull the slot machine lever once, each wheel spins independently. This means the appearance of an apple on one wheel does not influence whether another wheel shows an apple.
  • With independent events, you can multiply their probabilities to find the combined probability.
For instance, if the probability of a wheel not showing an apple in one spin is \( \frac{4}{5} \), then the probability of all five wheels not showing an apple in a single spin is: \[ \left( \frac{4}{5} \right)^5 = \frac{1024}{3125} \].

This calculation holds because the outcome of each wheel is not influenced by the others, highlighting the principle of independent events.
Equally Likely Outcomes
In probability, Equally Likely Outcomes means each outcome of a set has the same chance of occurring. This is one of the foundational assumptions in calculating probabilities.

In our slot machine example, each of the five symbols (apple, grape, peach, pear, and plum) appearing on a single wheel is equally likely. This implies each outcome has a probability of \( \frac{1}{5} \).
  • Understanding this helps us easily determine the likelihood of an event occurring.
For instance, the probability of not getting an apple is the sum of the probabilities of the other four symbols: \( \frac{4}{5} \).

This equal likelihood is a key assumption that allows us to calculate all other probabilities in this problem accurately. If the slots weren't equally likely to show any of the five symbols, we'd need additional information to proceed.
Calculation Steps
Grasping the Calculation Steps is paramount to solving probability problems effectively. Let鈥檚 break down how you approach the slot machine question.

Firstly, start by defining the probability of not getting an apple on a single wheel, \( P(\text{not apple}) = \frac{4}{5} \). Then, calculate the chance of no apples appearing on all wheels in a single spin. Since each wheel is independent, multiply these probabilities: \[ \left( \frac{4}{5} \right)^5 \]. This equals \( \frac{1024}{3125} \).
  • This result shows the probability of getting no apples in one single spin.
If you鈥檙e interested in acquiring at least one apple, employ the complement rule: subtract the no-apples probability from 1: \[ 1 - \frac{1024}{3125} = \frac{2101}{3125} \].

Finally, when considering multiple pulls of the lever, you compute the chance of no apples over those spins, and apply the complement again to find at least one apple鈥檚 probability over those spins.
By following these calculation steps, you can reliably solve for the likelihood of an event occurring, be it spinning apples or any other probabilistic scenario.

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

Do you Twitter? In late 2009, the Pew Internet and American Life Project asked a random sample of U.S. adults, 鈥淒o you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?鈥 According to Pew, the resulting 95% confidence interval is (0.167, 0.213). Can we use this interval to conclude that the actual proportion of U.S. adults who would say they Twitter differs from 0.20? Justify your answer.

You are testing \(H_{0} : \mu=10\) against \(H_{a} : \mu \neq 10\) based on an SRS of 15 observations from a Normal population. What values of the \(t\) statistic are statistically significant at the \(\alpha=0.005\) level? $$ \begin{array}{ll}{\text { (a) } t>3.326} & {\text { (d) } t<-3.326 \text { or } t>3.326} \\ {\text { (b) } t>3.286} & {\text { (e) } t<-3.286 \text { or } t>3.286}\end{array} $$ (c) \(t > 2.977\)

Which of the following 95\(\%\) confidence intervals would lead us to reiect \(H_{0} : p=0.30\) in favor of \(H_{a} : p \neq 0.30\) at the 5\(\%\) significance level? $$ \begin{array}{l}{\text { (a) }(0.29,0.38) \quad \text { (c) }(0.27,0.31) \quad \text { (e) None of these }} \\ {\text { (b) }(0.19,0.27) \quad \text { (d) }(0.24,0.30)}\end{array} $$

One-sided test Suppose you carry out a significance test of \(H_{0} : \mu=5\) versus \(H_{a} : \mu>5\) based on a sample of size \(n=20\) and obtain \(t=1.81\) (a) Find the P-value for this test using (i) Table B and (ii) your calculator. What conclusion would you draw at the 5% significance level? At the 1% significance level? (b) Redo part (a) using an alternative hypothesis of \(H_{a} : \mu \neq 5\)

Packaging \(\operatorname{CDs}(6.2,5.3)\) A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter \(\mu=4.2\) inches and standard deviation \(\sigma=0.05\) inches. The CD manufacturer decides to produce CDs with mean diameter \(\mu=4\) inches. Their diameters follow a Normal distribution with \(\sigma=0.1\) inches. (a) Let X the diameter of a randomly selected CD and Y the diameter of a randomly selected case. Describe the shape, center, and spread of the distribution of the random variable X Y. What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases.

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.