/*! 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 36 "N.Y. Lottery Numbers Come Up \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

"N.Y. Lottery Numbers Come Up \(9-1-1\) on \(9 / 11\) " was the headline of an article that appeared in the San Francisco Chronicle (September 13,2002 ). More than 5,600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. A professor at the University of Buffalo was quoted as saying, "I'm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up. People tend to read into these things. I'm sure that whatever numbers come up tonight, they will have some special meaning to someone, somewhere." The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. One ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) would be selected on any particular day?

Short Answer

Expert verified
The probability that the sequence \(9-1-1\) would be selected on any particular day is 0.001.

Step by step solution

01

Find the Probability of Drawing Each Number

Since each bin contains 10 balls (numbered 0-9), the probability of drawing a specific number from a single bin is \(\frac{1}{10}\) or 0.1.
02

Find the Probability of Drawing the Sequence \(9-1-1\)

As each draw is independent, the probability of drawing the sequence \(9-1-1\) would be the product of the probabilities of drawing a \(9\), then a \(1\), then another \(1\). Therefore, the probability is \(0.1 \times 0.1 \times 0.1 = 0.001\).

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
Understanding the concept of independent events is crucial when dealing with probability. In probability theory, two events are said to be independent if the occurrence of one event does not affect the probability of the occurrence of the other. This means that the outcome of one event has no bearing on the outcome of the second event.

For instance, consider rolling a die and flipping a coin. The result of the die roll does not influence whether the coin will land on heads or tails—they are independent events. Similarly, in the lottery example, the selection of each number is an independent event. The chance of drawing a '9' from the first bin is not impacted by the draw in the second or third bin.

This property allows us to calculate the combined probability of multiple independent events occurring by multiplying their individual probabilities. So, the probability of our desired sequence '9-1-1' happening is simply the product of the probabilities of each number being drawn independently, resulting in \(0.1 \times 0.1 \times 0.1 = 0.001\).
Probability Theory
At the heart of understanding random events is probability theory, which is a branch of mathematics concerned with the analysis of random phenomena. The fundamental component of probability is the 'event', which is any particular outcome or group of outcomes. Probability theory assigns a numerical value, ranging from 0 to 1, to the likelihood of an event occurring.

A probability of 0 indicates an impossibility, whereas a probability of 1 signifies certainty. Any probability in between reflects the degree of certainty that the event will happen. In our lottery problem, the probability of any specific number being drawn from one of the bins is \(\frac{1}{10}\) or 0.1, implying that each number has an equal chance to occur, reflecting the concept of a uniform distribution in this straightforward scenario.

The calculation of complex probabilities relies on rules and theorems developed within probability theory, such as the multiplication rule for independent events, which was utilized in computing the probability of the '9-1-1' sequence.
Randomness in Probability
Randomness is a key concept in probability that refers to the unpredictability and lack of pattern in events. It ensures that each event is governed by chance, with no discernible influence from prior outcomes or external factors.

In the context of the New York state lottery, the random selection of balls mimics the idea of randomness. Because each ball is equally likely to be chosen, every number combination has the same probability of being drawn, demonstrating the concept of a random process.

However, humans often look for patterns and may mistakenly see significance in sequences that are merely the result of random chance, such as the '9-1-1' sequence occurring on September 11th. This tendency is known as apophenia or patternicity, but it does not reflect the underlying principles of randomness where every number has an equal chance to be chosen, regardless of the sequence or date.

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

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a time line is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is 0.8 2\. If Maria completes her part on time, the probability that Alex completes on time is \(0.9,\) but if Maria is late, the probability that Alex completes on time is only 0.6 . 3\. If Alex completes his part on time, the probability that Juan completes on time is \(0.8,\) but if \(\mathrm{Alex}\) is late, the probability that Juan completes on time is only 0.5 . 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(0.9,\) but if Juan is late, the probability that Jacob completes on time is only 0.7 . Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time, and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

A large cable TV company reports the following: \- \(80 \%\) of its customers subscribe to its cable TV service \- \(42 \%\) of its customers subscribe to its Internet service \- \(32 \%\) of its customers subscribe to its telephone service \(25 \%\) of its customers subscribe to both its cable TV and Internet service \(21 \%\) of its customers subscribe to both its cable TV and phone service \- \(23 \%\) of its customers subscribe to both its Internet and phone service \- \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. Find and interpret the following probabilities: a. \(P(\) cable TV only \()\) b. \(P(\) Internet \(\mid\) cable \(\mathrm{TV})\) c. \(P\) (exactly two services) d. \(P\) (Internet and cable TV only)

The student council for a school of science and math has one representative from each of five academic departments: Biology (B), Chemistry (C), Mathematics (M), Physics (P), and Statistics (S). Two of these students are to be randomly selected for inclusion on a university-wide student committee. a. What are the 10 possible outcomes? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each outcome? c. What is the probability that one of the committee members is the statistics department representative? d. What is the probability that both committee members come from laboratory science departments?

In an article that appears on the website of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in \(100 .\) And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 - are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty \(T C=\) event that the test result is clean \(D=\) event that the person tested is actually dirty \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of i. \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a "hypothetical 1000 " table. c. What is the value of \(P(T D)\) ? d. Use the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

The Australian newspaper The Mercury (May 30,1995\()\) reported that based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who had quit smoking without a nicotine aid began smoking again within 2 weeks and \(60 \%\) of those who had used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who has attempted to quit begins smoking again within 2 weeks independent or dependent events? Justify your answer using the given information.

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.