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Consider the following two lottery-type games: Game 1: You pick one number between 1 and 50 . After you have made your choice, a number between 1 and 50 is selected at random. If the selected number matches the number you picked, you win. Game 2: You pick two numbers between 1 and 10 . After you have made your choices, two different numbers between 1 and 10 are selected at random. If the selected numbers match the two you picked, you win. a. The cost to play either game is \(\$ 1,\) and if you win you will be paid \(\$ 20 .\) If you can only play one of these games, which game would you pick and why? Use relevant probabilities to justify your choice. b. For either of these games, if you plan to play the game 100 times, would you expect to win money or lose money overall? Explain.

Short Answer

Expert verified
Based on the calculated probabilities and expected values, the better game to play would be Game 1. Despite the expected monetary loss in both games, Game 1 loses less money per game played on average compared to Game 2. Thus, one can expect to lose less money overall in Game 1 when played 100 times.

Step by step solution

01

Calculate Probability of Winning for Game 1

The probability of winning in Game 1 can be calculated using the formula for counting principle: \( P = 1/N \) where N is the total number of possibilities. Since there are 50 possible numbers that could be picked, the probability of winning Game 1 is \( 1/50 = 0.02 \)
02

Calculate Probability of Winning for Game 2

The probability of winning Game 2 is a bit more complicated since we're dealing with two numbers. We can calculate it by considering the two numbers are chosen independently, requiring us to use the product rule. The first number has 1 in 10 chance, and the second number (which must be different) has 1 in 9 chance. Hence, the probability of winning Game 2 is \( (1/10) * (1/9) = 0.01111 \)
03

Compare Probabilities

Comparing the probabilities calculated, Game 1 has a higher probability of winning at 0.02 compared to Game 2 with a winning probability of 0.01111. Therefore, based on probability of winning, Game 1 would be the better choice.
04

Calculate Expected Values

The expected value is the average amount of money we expect to win (or lose) per game played. For Game 1, we calculate the expected value by multiplying the probability of winning by the prize amount, and then subtracting the product of the probability of losing and the cost to play. Following this formula, the expected value of Game 1 is \( (0.02 * 20) - ((1 - 0.02) * 1) = -0.6 \). Similarly, for Game 2, the expected value is \( (0.01111 * 20) - ((1 - 0.01111) * 1) = -0.778 \)
05

Compare Expected Values

The game with the higher expected value is the more worthwhile to play. Here, Game 1 has the higher expected value at -0.6 compared to Game 2 with an expected value of -0.778. Thus, it is expected that for both games, the player would lose money, but they would lose less in Game 1. Hence, if the player plans to play the game 100 times, they are expected to lose more money in Game 2 than in Game 1.

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Key Concepts

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

Counting Principle
The counting principle is a fundamental concept in probability that allows us to determine the number of possible outcomes in a scenario where events occur in sequence and independently of one another. For example, let's say you wish to determine how many different two-letter sequences can be formed from the letters A, B, and C. To find this, you would use the counting principle which states that if there are 3 choices for the first letter and 3 choices for the second letter, then there are in total 3 times 3, which gives us 9 possible sequences.

When applied to games like lotteries or raffles, the counting principle tells us how many outcomes exist. For instance, if there is a game where you pick one number from 1 to 50, this means there are precisely 50 possible numbers you could pick, thus 50 possible outcomes. This straightforward principle is the foundation for calculating the probability of a single event with equally likely outcomes.
Product Rule in Probability
The product rule is crucial when dealing with independent events in probability. It states that the probability of two or more independent events occurring together is the product of their individual probabilities. What does this mean in practice? Let's look at a simple example involving dice. If you want to know the probability of rolling a four and then a six on a six-sided die, you would calculate the probability of rolling a four (1/6) and the probability of rolling a six (1/6) and multiply them together, thus \( (1/6) \times (1/6) = 1/36 \).

In the context of the lottery-type game where you pick two numbers out of 10, we acknowledge that each number is chosen independently. This is why for the second number, which has to be different from the first, we have 9 options instead of 10. We apply the product rule to find the overall probability of winning, thus \( \frac{1}{10} \times \frac{1}{9} = \frac{1}{90} \) or approximately 0.01111. Understanding this rule is very important for calculating probabilities in multi-stage games or processes.
Expected Value
In probability and statistics, the expected value is the sum of all possible values each multiplied by its probability of occurrence. This might sound complex, but it's actually a very practical concept. It basically tells us what we can expect to happen on average if we were to repeat an experiment (or game) over and over again.

How do we calculate expected value in real terms? For example, in a game where you bet \$1 to win \$20 with a probability of 1/50, the expected value is \( (1/50) \times 20 - (49/50) \times 1 \). This gives us a number that represents our average gains or losses per play. If this number is negative, it suggests that on average, we will lose money. It's a key indicator for making informed decisions about whether participating in a game or gamble is a financially sound choice.
Comparing Probabilities
When presented with multiple options, like different games in a casino, we often want to determine which choice gives us the best chance of winning. To do that, we compare their probabilities. This goes beyond just looking at the numbers; it involves understanding what these probabilities mean in the context of each game.

Using the two games from our example, we calculated a higher probability of winning in Game 1 than in Game 2. This, however, is not the end of the story. Probabilities also need to be weighed against potential payouts and the cost to play. For instance, a game with a lower probability of winning but a significantly higher payout might be more attractive than one with a higher probability but lower payout. When comparing probabilities, consider both the likelihood of winning and the potential rewards to decide which game offers the most value.

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Most popular questions from this chapter

A student placement center has requests from five students for employment interviews. Three of these students are math majors, and the other two students are statistics majors. Unfortunately, the interviewer has time to talk to only two of the students. These two will be randomly selected from among the five. a. What is the sample space for the chance experiment of selecting two students at random? (Hint: You can think of the students as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two students is \(\mathrm{A}\) and \(\mathrm{B}\). There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both selected students are statistics majors? d. What is the probability that both students are math majors? e. What is the probability that at least one of the students selected is a statistics major? f. What is the probability that the selected students have different majors?

A large cable company reports that \(80 \%\) of its customers subscribe to its cable TV service, \(42 \%\) subscribe to its Internet service, and \(97 \%\) subscribe to at least one of these two services. (Hint: See Example 5.6\()\) a. Use the given probability information to set up a "hypothetical \(1000 "\) table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer subscribes to both cable TV and Internet service. ii. the probability that a randomly selected customer subscribes to exactly one of these services.

Consider the following events: \(C=\) event that a randomly selected driver is observed to be using a cell phone \(A=\) event that a randomly selected driver is observed driving a car \(V=\) event that a randomly selected driver is observed driving a van or SUV \(T=\) event that a randomly selected driver is observed driving a pickup truck Based on the article "Three Percent of Drivers on Hand-Held Cell Phones at Any Given Time" (San Luis Obispo Tribune, July 24,2001 ), the following probability estimates are reasonable: \(P(C)=0.03, P(C \mid A)=0.026, P(C \mid V)=0.048\) and \(P(C \mid T)=0.019 .\) Explain why \(P(C)\) is not just the average of the three given conditional probabilities.

Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F)\) ? b. Suppose that for events \(A\) and \(B, P(A)=0.24, P(B)=0.24\), and \(P(A \cup B)=0.48 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

A study of how people are using online services for medical consulting is described in the paper "Internet Based Consultation to Transfer Knowledge for Patients Requiring Specialized Care" (British Medical Journal [2003]: \(696-699)\). Patients using a particular online site could request one or both (or neither) of two services: specialist opinion and assessment of pathology results. The paper reported that \(98.7 \%\) of those using the service requested a specialist opinion, \(35.4 \%\) requested the assessment of pathology results, and \(34.7 \%\) requested both a specialist opinion and assessment of pathology results. Consider the following two events: \(S=\) event that a specialist opinion is requested \(A=\) event that an assessment of pathology results is requested a. What are the values of \(P(S), P(A)\), and \(P(S \cap A)\) ? b. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(S\) and not \(S\) and rows corresponding to \(A\) and \(\operatorname{not} A .\) c. Use the table to find the following probabilities: i. the probability that a request is for neither a specialist opinion nor assessment of pathology results. ii. the probability that a request is for a specialist opinion or an assessment of pathology results.

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