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Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution. \(\bullet\) Let \(X=\) the number of years a student will study ballet with the teacher. \(\bullet\) Let \(P(x)=\) the probability that a student will study ballet \(x\) years. You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win \(\$ 30\). If it is not a face card, you pay \(\$ 2\). There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?

Short Answer

Expert verified
The expected value of the game is approximately \( \$5.39 \).

Step by step solution

01

Determine Probabilities

We first identify the probability of drawing a face card and not drawing a face card from the deck. There are 12 face cards in a 52-card deck, so the probability of drawing a face card is \( P(\text{face card}) = \frac{12}{52} = \frac{3}{13} \). Consequently, the probability of not drawing a face card is \( P(\text{not a face card}) = 1 - \frac{3}{13} = \frac{40}{52} = \frac{10}{13} \).
02

Determine Values of Outcomes

Next, we establish the monetary outcomes for the game. Winning results in a gain of \( \\(30 \) for drawing a face card, whereas losing results in a loss of \( \\)2 \) for not drawing a face card.
03

Calculate Expected Value

To find the expected value, we use the formula for expected value \( E(X) = \sum [x_i \times P(x_i)] \). This gives us: \( E(X) = (30 \times \frac{3}{13}) + (-2 \times \frac{10}{13}) \). Calculating the first term: \( 30 \times \frac{3}{13} = \frac{90}{13} \approx 6.923 \). For the second term: \(-2 \times \frac{10}{13} = -\frac{20}{13} \approx -1.538 \).
04

Sum the Expected Values

Add the results from the outcomes: \( 6.923 + (-1.538) = 5.385 \). Thus, the expected value of the game is approximately \( \$5.39 \).

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Key Concepts

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

Expected Value
The expected value is a key concept in probability theory. It provides a measure of the central tendency of a random variable. To put it simply, the expected value predicts the average outcome of an event if you were to repeat the event multiple times.
For example, in our card game scenario, we are looking to find out what we can expect to gain or lose if we play this game over a long period. Using the expected value helps us determine this.
To calculate the expected value in this context, we use the formula: \[ E(X) = \sum [x_i \times P(x_i)] \] Where \( x_i \) represents the outcome and \( P(x_i) \) is the probability of that outcome.
The formula involves multiplying each possible outcome by its corresponding probability and summing up all these products. - In our exercise: - Winning \\(30 happens with a probability of \( \frac{3}{13} \) - Losing \\)2 happens with a probability of \( \frac{10}{13} \) - Hence, the expected value is \( (30 \times \frac{3}{13}) + (-2 \times \frac{10}{13}) \).By computing these, we find that playing the game has an expected gain of about \\(5.39 per round. This means if you play the game many times, you could expect to win an average of \\)5.39 per game.
Probability Distribution
Probability distribution is a concept that provides a complete description of the probabilities of all possible outcomes of a random variable. In our initial exercise, the probability distribution helps in understanding how long students tend to study ballet over the years with a particular teacher.
Probability distribution can take many forms, but for our card game problem, we are mainly interested in the discrete probability distribution for two outcomes: drawing a face card or not.
- Let's look at it: - Probability of drawing a face card: \( P(\text{face card}) = \frac{3}{13} \) - Probability of not drawing a face card: \( P(\text{not a face card}) = \frac{10}{13} \)A table isn't needed to comprehend these simple outcomes. Each outcome has its probability, creating a probability distribution that covers every possible outcome.
In essence, knowing how probability distribution works helps inform decisions such as choosing classes to offer or predicting winnings in games. Understanding them allows you to interpret how likely certain events are relative to others, thereby aiding in effective planning and decision-making.
Probability Calculations
Probability calculations form the core of probability theory by allowing us to determine the likelihood of various outcomes. They help assess risks and gain insights into future events based on current data.
  • **Step 1: Identify Probabilities** - For our card game, knowing the proportions of face cards and non-face cards in a deck (\(\frac{3}{13}\) and \(\frac{10}{13}\) respectively) was our first step.

  • **Step 2: Determine Outcomes** - Here, monetary gains or losses associated with drawing a face or non-face card were calculated. Winning yields \\(30 and losing requires paying \\)2.
By combining these probabilities and outcomes in the expected value formula, we computed the game's average outcome. Practicing these steps enhances our ability to deal with uncertain events, equipping us with the knowledge to predict potential results accurately. Understanding probability calculations can navigate everyday uncertainties, making it an invaluable skill in various aspects of life and academics.

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Most popular questions from this chapter

Use the following information to answer the next seven exercises: A ballet instructor is interested in knowing what percent of each year's class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution. \(\bullet\) Let \(X=\) the number of years a student will study ballet with the teacher. \(\bullet\) Let \(P(x)=\) the probability that a student will study ballet \(x\) years. What does the column \(" P(x)\) "sum to and why?

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