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Calculate the expected value of the given random variable \(X .\) [Exercises \(23,24,27\), and 28 assume familiarity with counting arguments and probability (Section 7.4).] \(\nabla 30\) darts are thrown at a dartboard. The probability of hitting a bull's-eye is \(\frac{1}{5}\). Let \(X\) be the number of bull's-eyes hit.

Short Answer

Expert verified
The expected value of the random variable $X$, representing the number of bull's-eyes hit when throwing 30 darts with a probability of $\frac{1}{5}$ for each dart to hit a bull's-eye, can be calculated using the formula $E(X) = np$. In this case, $n=30$ and $p=\frac{1}{5}$, so the expected value is $E(X) = (30) (\frac{1}{5}) = 6$.

Step by step solution

01

Applying the binomial distribution

We know that each dart has a probability of 1/5 of hitting a bull's-eye and the total number of darts thrown is 30. We can model this situation using the binomial probability formula: \[ P(X=k) = {n\choose k} p^k (1-p)^{n-k} \] where n represents the number of trials (30 darts), k represents the number of successes (bull's-eyes hit), p represents the probability of success (1/5 chance of hitting a bull's-eye), and \({n\choose k}\) is the number of possible combinations of k successes out of n trials.
02

Calculating the expected value using binomial probabilities

Using the formula for the expected value of a discrete random variable, we can find the expected value by inserting the probabilities determined in Step 1: \[ E(X) = \sum_{k=0}^{30} k {30\choose k} (\frac{1}{5})^k (\frac{4}{5})^{30-k} \] However, to make calculations simpler, we can use the alternative formula for the expected value of a binomial distribution: \[ E(X) = np \]
03

Calculating the expected value using the simplified formula

Now, we can simply plug in the total number of darts (30) and the probability of hitting a bull's-eye (1/5) to calculate the expected value: \[ E(X) = (30) (\frac{1}{5}) \] \[ E(X) = 6 \] The expected value of the random variable X, which represents the number of bull's-eyes hit when throwing 30 darts with a probability of 1/5 for each dart to hit a bull's-eye, is 6.

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Key Concepts

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

Expected Value
In probability theory, the expected value is a fundamental concept that represents the "center" or "average" of a random variable's possible outcomes. Imagine you are playing a game that involves a certain level of chance. The expected value helps you predict, on average, what you are likely to gain or lose over many iterations of the game.

For a discrete random variable like the number of bull's-eyes hit with darts, the expected value is calculated by multiplying each possible outcome by its probability and then summing all these products. This gives us a balanced average. Fortunately, when dealing with binomial distributions, we have a shortcut. The formula simplifies to \( E(X) = np \). Here, \( n \) is the total number of trials (darts thrown, in our example), and \( p \) is the probability of success on each trial (hitting a bull's-eye).

Using this method, the expected value is straightforward to calculate. It becomes especially useful in predicting outcomes in various fields, from gambling to stock market investments.
Probability Theory
Probability theory is the branch of mathematics that deals with uncertainty. It allows us to quantify the likelihood of different outcomes. For instance, when throwing darts, we aren't certain how many bull's-eyes we will hit, but probability theory lets us calculate the expected number of hits.

We often use models like the binomial distribution to represent these kinds of scenarios. In the case of the darts example, each dart throw can be viewed as an independent trial with two possible outcomes (hit or miss). The binomial distribution is apt for modeling such scenarios because it considers the number of successes (bull's-eyes) across a fixed number of independent trials (darts).

The probability of each outcome can be determined using the binomial probability formula \( P(X=k) = {n \choose k} p^k (1-p)^{n-k} \). This formula helps in understanding the likelihood of, say, hitting a certain number of bull's-eyes out of 30 darts, when the probability of a bull's-eye is 1/5 per dart.
  • Probabilities range from 0 (impossible event) to 1 (certain event).
  • They provide a foundation for further statistical analysis and decision-making.
Discrete Random Variables
A discrete random variable is one that can take on a countable number of distinct values. In our darts example, the random variable \( X \) represents the number of bull's-eyes achieved, which can be any integer between 0 and 30.

Discrete random variables are characterized by their probability distribution, which lists each possible outcome and the probability that it will occur. For instance, with 30 dart throws where each has a 1/5 chance of hitting a bull's-eye, the random variable \( X \) might look something like a list of values, each paired with its probability.

This concept is critical because it allows us to design and analyze experiments. From the probability distribution, we can derive important measures such as the expected value and variance, which help in understanding the random variable's behavior.
  • Discrete variables have finite or countably infinite possible values.
  • They are key in statistical analyses and probabilities.

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Most popular questions from this chapter

Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{20}\) $$ -100,400,-200,-500,200,-300,-200,900,-100,200 $$ Compute the mean and median of the given sample. Fill in the blank: There were as many days with a change in the Dow above \(\quad\) points as there were with changes below that.

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The new computer your business bought lists a mean time between failures of 1 year, with a standard deviation of 2 months. Ten months after a repair, it breaks down again. Is this surprising? (Assume that the times between failures are normally distributed.)

The Blue Sky Flight Insurance Company insures passengers against air disasters, charging a prospective passenger \(\$ 20\) for coverage on a single plane ride. In the event of a fatal air disaster, it pays out \(\$ 100,000\) to the named beneficiary. In the event of a nonfatal disaster, it pays out an average of \(\$ 25,000\) for hospital expenses. Given that the probability of a plane's crashing on a single trip is \(.00000087,{ }^{32}\) and that a passenger involved in a plane crash has a \(.9\) chance of being killed, determine the profit (or loss) per passenger that the insurance company expects to make on each trip. HINT [Use a tree to compute the probabilities of the various outcomes.]

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