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Chapter 7: Problem 32

In "The Numbers Game," a state lottery, four numbers are drawn with replacement from an urn containing balls numbered \(0-9\), inclusive. Find the probability that a ticket holder has the indicated winning ticket. Two specified, consecutive digits in exact order (the first two digits, the middle two digits, or the last two digits)

Short Answer

Expert verified
The probability of having a winning ticket with two specified consecutive digits in exact order (the first two digits, the middle two digits, or the last two digits) in "The Numbers Game" is \(0.2684\) or \(26.84\%\).

Step by step solution

01

Calculate total possible outcomes

Since the lottery involves drawing four numbers with replacement from an urn containing ten balls, each draw has a possibility of ten outcomes. Therefore, the total possible outcomes are represented by 10^4. Total possible outcomes = \(10^4 = 10000\)
02

Calculate possible outcomes for winning tickets using the specified rules

Now, we need to find the total successful winning ticket outcomes for each of the conditions, which are: 1. First two digits are exact consecutive numbers 2. Middle two digits are exact consecutive numbers 3. Last two digits are exact consecutive numbers For each condition, the given digits are specified and, therefore, fixed while the other two draws maintain their 10 possibilities. But, we have to consider the fact that we should not double count cases where the two specific consecutive digits overlap in two different scenarios. If the first two digits are fixed: There are 9 possibilities (either 01, 12, 23, 34, 45, 56, 67, 78, or 89) for the sequence, and 10 possibilities for each of the other two draws. Number of possible winning tickets for condition 1 = \(9 * 10^2 = 900\) If the middle two digits are fixed: Similar to condition 1, we have 9 possibilities for the fixed digits and 10 possibilities for the first and last draws. Number of possible winning tickets for condition 2 = \(9 * 10^2 = 900\) If the last two digits are fixed: Similar to condition 1 & 2, we have 9 possibilities for the fixed digits and 10 possibilities for the first two draws. Number of possible winning tickets for condition 3 = \(9 * 10^2 = 900\) However, as previously mentioned, there are cases that overlap. We have to subtract these overlapping cases between condition 1 and 2, condition 2 and 3, and condition 1 and 3. Overlap between 1 and 2: Middle two digits are fixed, and last three digits form a consecutive ascending sequence. There are 8 possibilities (012, 123, 234, 345, 456, 567, 678, or 789). Overlap between 2 and 3: First three digits form a consecutive ascending sequence. There are again 8 possibilities (012, 123, 234, 345, 456, 567, 678, or 789). Overlap between 1 and 3: Impossible, since 4 numbers cannot form two separate consecutive sequences. Hence, we have to subtract twice the overlapping cases (16) from the sum of the successful outcomes. Total successful outcomes = (900 + 900 + 900) - 16 = 2684
03

Calculate the probability of having a winning ticket

Finally, to find the probability, divide the total successful outcomes by the total possible outcomes. Probability = \(\frac{2684}{10000} = 0.2684\) So, the probability of having a winning ticket with the given conditions is 0.2684 or 26.84%.

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Key Concepts

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

Finite Mathematics
Finite mathematics is an area of mathematics that deals with mathematical concepts and techniques that are used in real-world applications and fields such as business, engineering, the social sciences, and the life sciences. Unlike calculus, which deals with continuous variables and can have an infinite number of possible solutions, finite mathematics focuses on discrete elements that have a limited set of possible outcomes.

In the context of our lottery problem, finite mathematics comes into play through the use of combinatorics and probability calculation; specifically, determining the finite number of outcomes when four numbers are drawn with replacement. We observe that the set of numbers in the urn is finite (only numbers 0-9) and, therefore, the possible sequences that can be drawn are also finite. This finite nature allows us to use specific mathematical methods to accurately calculate probabilities.
Combinatorics
Combinatorics is a branch of mathematics that studies the counting, arrangement, and combination of elements within a set. It's fundamental to the study of probability because it provides tools to count the number of possible outcomes and configurations without actually listing them all.

In lotteries, combinatorics would help us understand how many different ways we can select our winning numbers. For example, with ten different numbered balls and the chance to draw four numbers, we can use combinatorial principles to calculate the total number of different sequences possible鈥攖hat's where the calculation of '10 to the power of 4' comes from. Additionally, combinatorics can help address more complex problems, such as the number of sequences with specified restrictions, like consecutive numbers, which is crucial for our exercise.
Probability Calculation
Probability calculation is the process of determining the likelihood of an event occurring. It is a ratio of the number of favorable outcomes to the total number of possible outcomes and is often expressed as a number between 0 and 1, or as a percentage.

In the lottery problem, the probability calculation helps us understand the chances of drawing a winning ticket with particular constraints. By first identifying the total possible outcomes and then the number of successful outcomes for each condition鈥攖aking care not to double-count any overlapping scenarios鈥攚e can determine the probability of winning. This calculation is foundational in probability theory and helps in making predictions about the outcomes of random events.
Replacement in Probability
Replacement in probability refers to a scenario where an element is returned to a set after being drawn, meaning that it could potentially be selected again in the same series of events. This concept affects how we calculate probabilities because the probability of selecting each element remains constant from one draw to another.

In our lottery problem, numbers are drawn with replacement, which means that after a number is drawn, it is placed back into the urn before the next draw. This replacement ensures that each draw is independent of the previous ones. The total number of outcomes thus remains constant at 10 for each draw. However, when dealing with specific sequences like consecutive numbers, we also pay attention to the fact that certain sequences can overlap across different conditions鈥攚hich slightly alters the simple application of replacement. This consideration is a key factor in correctly calculating the odds of winning in certain types of lottery draws.

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Most popular questions from this chapter

A medical test has been designed to detect the presence of a certain disease. Among those who have the disease, the probability that the disease will be detected by the test is \(.95\). However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is .04. It is estimated that \(4 \%\) of the population who take this test have the disease. a. If the test administered to an individual is positive, what is the probability that the person actually has the disease? b. If an individual takes the test twice and both times the test is positive, what is the probability that the person actually has the disease? (Assume that the tests are independent.)

In an attempt to study the leading causes of airline crashes, the following data were compiled from records of airline crashes from 1959 to 1994 (excluding sabotage and military action). $$\begin{array}{lc} \hline \text { Primary Factor } & \text { Accidents } \\ \hline \text { Pilot } & 327 \\ \hline \text { Airplane } & 49 \\ \hline \text { Maintenance } & 14 \\ \hline \text { Weather } & 22 \\ \hline \text { Airport/air traffic control } & 19 \\ \hline \text { Miscellaneous/other } & 15 \\ \hline \end{array}$$ Assume that you have just learned of an airline crash and that the data give a generally good indication of the causes of airline crashes. Give an estimate of the probability that the primary cause of the crash was due to pilot error or bad weather.

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?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) and \(B\) are events of an experiment, then $$ P(A \cap B)=P(A \mid B) \cdot P(B)=P(B \mid A) \cdot P(A) $$

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) and \(B\) are mutually exclusive and \(P(B) \neq 0\), then \(P(A \mid B)=0 .\)
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