/*! 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 117 Probability played a role in the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Probability played a role in the rigging of the April \(24,1980,\) Pennsylvania state lottery. To determine each digit of the three-digit winning number, each of the numbers \(0,1,2, \ldots, 9\) is written on a Ping-Pong ball, the 10 balls are blown into a compartment, and the number selected for the digit is the one on the ball that floats to the top of the machine. To alter the odds, the conspirators injected a liquid into all balls used in the game except those numbered 4 and 6 , making it almost certain that the lighter balls would be selected and determine the digits in the winning number. They then proceeded to buy lottery tickets bearing the potential winning numbers. How many potential winning numbers were there (666 was the eventual winner)?

Short Answer

Expert verified
Answer: 8 potential winning numbers.

Step by step solution

01

Identify the possible digits

To find the potential winning numbers, we must first identify the possible digits that can be used to form the winning number. Since the balls with numbers 4 and 6 were left untouched, we can assume that these are the only digits we can use to form the winning number.
02

Calculate the total number of potential winning numbers

Now that we have identified the possible digits (4 and 6), we can proceed with generating the potential winning numbers. There are three positions in a 3-digit number, and for each position, we have two choices (4 or 6). Hence, we can calculate the total number of possible winning numbers using the multiplication principle: Total potential winning numbers = 2 * 2 * 2
03

Calculate the result

Multiplying the number of possibilities for each digit, we get: Total potential winning numbers = 2 * 2 * 2 = 8 So, there were 8 potential winning numbers.

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Key Concepts

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

Probability Principles
Probability principles are foundational guidelines for understanding how likely an event is to occur. In the context of lottery games, these principles can help predict the chance of winning. Probabilities are calculated by considering the total number of possible outcomes and the number of ways an event can occur. For a fair lottery game, where each combination of numbers has an equal chance of being drawn, the probability of winning is typically very low because the number of possible number combinations is quite large.

However, as shown in the Pennsylvania lottery rigging case, manipulating the physical properties of the balls changed the probability of certain numbers being drawn. The conspirators created a situation where only the numbers 4 and 6 had a usual probability to come up, turning the event into a non-random, almost certain outcome for these numbers. This significantly increased the odds in their favor and reduced the complexity of calculating the possible outcomes. Understanding how these alterations impact the probability is crucial in grasping how the integrity of a lottery can be compromised.
Combinatorics
Combinatorics is a branch of mathematics that deals with counting, combination, and permutation of sets of elements. In the realm of lottery games, combinatorics is utilized to determine the number of ways winning numbers can be selected. For example, in a standard lottery where you must choose 6 numbers out of 49, combinatorics would help to calculate the total number of possible combinations.

In our rigged lottery scenario, since only two numbers can be drawn to form the winning number (either 4 or 6), combinatorics becomes simplified to basic counting. The possible outcome for each of the digits that can be used to form the three-digit number is just 2 (either 4 or 6). Using combinatorial principles in this case, we would calculate the different arrangements that these limited choices could produce, demonstrating just how drastically the rigging influenced the number of possible outcomes and paving the way to a fraudulent win.
Multiplication Principle
The multiplication principle is a fundamental rule used in counting that allows us to compute the number of possible outcomes in a sequence of events. It states that if one event can occur in 'm' ways and another event can occur independently in 'n' ways, then the total number of ways both events can occur is 'm x n'.

In the example of the fixed lottery, the multiplication principle simplifies the process of finding the total potential winning numbers. There are three digits to be chosen, and each has 2 possibilities (4 or 6). Hence, for the first digit, we have 2 choices; for the second digit, again 2 choices; and the same for the third, leading to a calculation of the total outcomes as 2 times 2 times 2. This principle, when applied correctly, can reveal the extent of a game's complexity or simplicity, as it did in uncovering the limited outcomes in the Pennsylvania state lottery scenario.

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Most popular questions from this chapter

An experiment involves tossing a single die. These are some events: A: Observe a 2 \(B:\) Observe an even number \(C:\) Observe a number greater than 2 \(D:\) Observe both \(A\) and \(B\) \(E:\) Observe \(A\) or \(B\) or both \(F:\) Observe both \(A\) and \(C\) a. List the simple events in the sample space. b. List the simple events in each of the events \(A\) through \(F\) c. What probabilities should you assign to the simple events? d. Calculate the probabilities of the six events \(A\) through \(F\) by adding the appropriate simple-event probabilities.

Suppose that \(P(A)=.3\) and \(P(B)=.5\). If events \(A\) and \(B\) are mutually exclusive, find these probabilities: a. \(P(A \cap B)\) b. \(P(A \cup B)\)

An experiment can result in one of five equally likely simple events, \(E_{1}, E_{2}, \ldots, E_{5} .\) Events \(A, B,\) and \(C\) are defined as follows: $$A: E_{1}, E_{3} \quad P(A)=.4$$ $$B: E_{1}, E_{2}, E_{4}, E_{5} \quad P(B)=.8$$ $$C: E_{3}, E_{4} \quad P(C)=.4$$. Find the probabilities associated with these compound events by listing the simple events in each. a. \(A^{c}\) b. \(A \cap B\) c. \(B \cap C\) d. \(A \cup B\) e. \(B \mid C\) f. \(A \mid B\) g. \(A \cup B \cup C\) h. \((A \cap B)^{c}\)

You have two groups of distinctly different items, 10 in the first group and 8 in the second. If you select one item from each group, how many different pairs can you form?

A survey classified a large number of adults according to whether they were judged to need eyeglasses to correct their reading vision and whether they used eyeglasses when reading. The proportions falling into the four categories are shown in the table. (Note that a small proportion, .02, of adults used eyeglasses when in fact they were judged not to need them.) $$\begin{array}{lcc} & \begin{array}{l}\text { Used Eyeglasses } \\\\\text { for Reading }\end{array} \\\\\hline \text { Judged to Need } & & \\ \text { Eyeglasses } & \text { Yes } & \text { No } \\\\\hline \text { Yes } & .44 & .14 \\\\\text { No } & .02 & .40\end{array}$$ If a single adult is selected from this large group, find the probability of each event: a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not.

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