/*! 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 27 Suppose you play a carnival game... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose you play a carnival game that requires you to toss a ball to hit a target. The probability that you will hit the target on each play is .2 and is independent from one try to the next. You win a prize if you hit the target by the third try. a. What is the probability that you hit the target on the first try? b. What is the probability that you miss the target on the first try but hit it on the second try? c. What is the probability that you miss the target on the first and second tries but hit it on the third try? d. What is the probability that you win a prize?

Short Answer

Expert verified
a. 0.2, b. 0.16, c. 0.128, d. 0.488

Step by step solution

01

Probability of Hitting on the First Try

To find the probability of hitting the target on the first try, we simply use the given probability of hitting the target on any play. This probability is \(P( ext{Hit on 1st try}) = 0.2\).
02

Probability of Missing First but Hitting Second

To find this probability, we need to first miss the first try and then hit the second try. The probability of missing is \(1 - 0.2 = 0.8\). Thus, the probability of missing the first try and hitting the second try is:\[P( ext{Miss 1st, Hit 2nd}) = 0.8 \times 0.2 = 0.16\]
03

Probability of Missing First and Second but Hitting Third

For this scenario, we must miss the first try, miss the second try, and hit on the third try. This probability is:\[P( ext{Miss 1st, Miss 2nd, Hit 3rd}) = 0.8 \times 0.8 \times 0.2 = 0.128\]
04

Probability of Winning a Prize

To win a prize, you must hit the target on either the first, second, or third tries. The probability of hitting the target in any of these tries corresponds to the sum of the probabilities from Step 1, Step 2, and Step 3:\[P( ext{Win a Prize}) = 0.2 + 0.16 + 0.128 = 0.488\]

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.

Independent Events
In probability, independent events refer to scenarios where the occurrence of one event does not affect the probability of the other event occurring. This can be thought of like flipping a coin or rolling a dice multiple times, where past outcomes do not influence future ones. In the carnival game described, each ball toss is an independent event.

This means that hitting the target or missing it on one throw does not change the likelihood of hitting or missing on the next throw. The probability remains the same for each independent event. They are unaffected by previous outcomes. This concept is crucial because it simplifies the calculation of probabilities over multiple attempts.

When calculating probabilities for sequences of independent events, such as missing the first and hitting the second, it allows us to simply multiply the probabilities of each individual event happening. For example:
  • The probability of missing the target on the first and then hitting on the second toss: The necessary operation is multiplying separate probabilities: \[0.8 \times 0.2 = 0.16\]
Probability of Success
The term 'probability of success' is often used in probability theory and statistics to denote the likelihood of achieving a predefined successful outcome. In this exercise, hitting the target can be considered a success.

Knowing the probability of success on a single try helps in determining the probabilities for different scenarios across multiple tries. For example, the probability of success for hitting the target on any given attempt is given as 0.2, or 20%. This is crucial because it forms the base for calculating more complex probabilities when multiple attempts are involved, such as finding the probability of hitting the target on at least one of several tries.

When calculating sequential tries, simply use individual success probabilities for each try. This way, the total likelihood of success over many attempts can be determined. Remember, the probability of success and the probability of failure (missing the target), which is calculated as \(1 - \text{probability of success}\), are complementary.
  • Probability of a successful hit on the first toss: \(P(\text{Hit on 1st try}) = 0.2\)
  • Probability of failing (or missing) on a single try: \(P(\text{Miss}) = 1 - 0.2 = 0.8\)
Cumulative Probability
Cumulative probability is the probability that a random variable takes on a value less than or equal to a specific value. In the context of the carnival game, this concept helps in calculating the probability of winning a prize up to and including the third try.

By calculating cumulative probability, we determine the likelihood of achieving at least one success in a series of attempts, i.e., hitting the target on the first, second, or third try. It aggregates the probabilities of all possible successful outcomes to provide a single probability value.

To find the cumulative probability of hitting the target and winning a prize by the third try, we add up the separate probabilities of all scenarios where you hit the target:
  • Probability of hitting on the first attempt: \(0.2\)
  • Probability of hitting on the second attempt after missing the first: \(0.16\)
  • Probability of hitting on the third attempt after missing the first two: \(0.128\)
  • Combined probability of at least one success up to three tries: \(0.2 + 0.16 + 0.128 = 0.488\)
This cumulative probability tells us there is a 48.8% chance of winning a prize if you have up to three tries.

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

A restaurant server knows that the probability that a customer will order coffee is .30, the probability that a customer will order a diet soda is. \(40,\) and the probability that a customer will request a glass of water is \(.70 .\) Explain what is wrong with his reasoning in each of the following. a. Using Rule 2 , he concludes that the probability that a customer will order either coffee or request a glass of water is \(.30+.70=1.0\) b. Using Rule 3, he concludes that the probability that a customer will order coffee and a diet soda is \((.30)(.40)=.12\)

Suppose you routinely check coin-return slots in vending machines to see if they have any money in them. You have found that about \(10 \%\) of the time you find money. a. What is the probability that you do not find money the next time you check? b. What is the probability that the next time you will find money is on the third try? c. What is the probability that you will have found money by the third try?

There is something wrong in each of the following statements. Explain what is wrong. a. The probability that a randomly selected driver will be wearing a seat belt is .75, whereas the probability that he or she will not be wearing one is .30. b. The probability that a randomly selected car is red is 1.20 . c. The probability that a randomly selected car is red is \(.20,\) whereas the probability that a randomly selected car is a red sports car is .25.

Use the probability rules in this chapter to solve each of the following: a. According to the U.S. Census Bureau, in 2012 , the probability that a randomly selected child in the United States was living with his or her mother as the sole parent was .244 and with his or her father as the sole parent was. \(040 .\) What was the probability that a child was living with just one parent? b. In 2010 in the United States, the probability that a birth would result in twins was \(.0331,\) and the probability that a birth would result in triplets or more was .0014. What was the probability that a birth in 2010 resulted in a single child?

We have seen many examples for which the term expected value seems to be a misnomer. Construct an example of a situation in which the term expected value would not seem to be a misnomer for what it represents.

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.