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Geometric or not? Determine whether each of the following scenarios describes a geometric setting. If so, define an appropriate geometric random variable. (a) Shuffle a standard deck of playing cards well. Then turn over one card at a time from the top of the deck until you get an ace. (b) Lawrence is learning to shoot a bow and arrow. On any shot, he has about a 10\(\%\) chance of hitting the bull's-eye. Lawrence's instructor makes him keep shooting until he gets a bull's-eye.

Short Answer

Expert verified
Both scenarios are geometric settings with defined random variables for each: cards drawn until an ace, and arrows shot until a bull's-eye.

Step by step solution

01

Understand Geometric Distribution

In a geometric setting, we are looking for scenarios where we perform a series of independent trials until we achieve the first 'success'. The trials are independent, and each trial has the same probability of success.
02

Analyze Scenario (a)

Scenario (a) deals with drawing cards from a shuffled deck until an ace is drawn. Each card draw is an independent event, with each card having an equal probability of being an ace. This fits the geometric model: "success" is drawing an ace.
03

Define the Geometric Random Variable for (a)

In scenario (a), define the geometric random variable as the number of cards drawn until the first ace appears. The probability of success (drawing an ace) in one draw is \( p = \frac{4}{52} = \frac{1}{13} \).
04

Analyze Scenario (b)

Scenario (b) involves Lawrence shooting arrows until he hits the bull's-eye. Each shot can be considered an independent trial, with a constant probability of hitting the bull's-eye (10\%). This matches the definition of a geometric setting, where each trial has probability \( p = 0.10 \) of being a "success".
05

Define the Geometric Random Variable for (b)

For scenario (b), define the geometric random variable as the number of shots taken until Lawrence hits the bull's-eye. The probability of success (hitting the bull's-eye) on any shot is \( p = 0.10 \).

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Key Concepts

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

Probability
Probability is the heart of understanding geometric distribution. It quantifies the likelihood of a particular outcome occurring in an experiment. In the context of geometric distribution, it specifically relates to the probability of success on a given trial.

Consider flipping a coin. The probability of landing on heads is 50%. This same concept applies to other scenarios, like drawing a card from a deck or shooting at a target. To fully grasp probability:
  • Identify the total number of possible outcomes.
  • Identify the number of favorable outcomes, or successes, specific to your interest.
  • Divide the number of successes by the total possible outcomes.
Understanding probability allows you to set the stage for applying the geometric model. It remains constant across all trials in a geometric distribution, which is a distinctive feature of this type of probability problem.

For example, in card-drawing scenarios, "success" might be drawing an ace. Here, the probability is calculated based on the number of aces in the deck compared to the total number of cards. Probability helps predict how soon a success is likely to occur in repeated trials.
Random Variables
Random variables are indispensable in probability theory, especially when dealing with geometric distributions. A random variable represents numerical outcomes associated with a probabilistic event.

There are different types of random variables, but in the geometric setting, we are usually concerned with discrete random variables. These denote counts, such as the number of attempts required to achieve the first success.

In scenarios involving geometric distributions, random variables are used to model the number of trials until the first success occurs. This is where the "geometric random variable" comes into play:
  • Defined as the number of independent trials taken to achieve the first success.
  • Each trial is identical, with a fixed probability of success.
  • The value of the random variable changes as the number of trials varies.
A clear example would be defining a random variable for how many arrows Lawrence needs to shoot to hit a bull's-eye. The randomness here lies in the outcome of each arrow attempt. Being able to define and manipulate random variables is crucial for calculating probabilities and expectations in the geometric distribution.
Independent Trials
In any geometric distribution scenario, the concept of independent trials is fundamental. Independent trials mean that the outcome of one trial does not affect the outcome of the other trials. Each trial in the sequence behaves as if it's the very first one, thereby ensuring true randomness.

To better understand independent trials, consider the examples described in the exercise:
  • Drawing cards from a deck: Each draw doesn't influence the probability of success on the next one because the card is replaced or because in this context, the knowledge of previous cards doesn't change the setup for subsequent ones (ideal condition).
  • Shooting arrows: Hitting or missing the bull's-eye on one shot doesn't change the probability associated with the following shots for Lawrence.
Independent trials allow us to multiply probabilities across trials when calculating the likelihood of a series of outcomes. They simplify complex problems by making each trial analyzable on its own.

Without the condition of independence, the geometric distribution would not apply, as the probabilities of each trial must remain constant to properly model the distribution. The sense of a fresh start on each trial is what preserves the integrity of the geometric setting.

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Most popular questions from this chapter

Geometric or not? Determine whether each of the following scenarios describes a gcometric setting. If so, define an appropriate gcometric random variable. (a) A popular brand of cereal puts a card with one of five famous NASCAR drivers in each box. There is a 1\(/ 5\) chance that any particular driver's card ends up in any box of cereal. Buy boxes of the cereal until you have all 5 drivers' cards. (b) In a game of \(4-\) Spot Keno, Lola picks 4 \(\mathrm{num}\)bers from 1 to \(80 .\) The casino randomly selects 20 winning numbers from 1 to \(80 .\) Lola wins money if she picks 2 or more of the winning numbers. The probability that this happens is \(0.259 .\) Lola decides to keep playing games of \(4-\) Spot Keno until she wins some money.

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A deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: Draw one card at random from the deck. You win \(\$ 10\) if the card drawn is an ace. Otherwise, you lose \(\$ 1 .\) If you make this wager very many times, what will be the mean amount you win? (a) About \(-\$ 1\) , because you will lose most of the time. (b) About \(\$ 9,\) because you win \(\$ 10\) but lose only \(\$ 1\) . (c) About \(-\$ 0.15 ;\) that is, on average you lose about 15 cents. (d) About \(\$ 0.77 ;\) that is, on average you win about 77 cents. (e) About \(\$ 0,\) because the random draw gives you a fair bet.

Checking for survey errors One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About 12\(\%\) of American adults identify themselves as black. Suppose we take an SRS of 1500 American adults and let \(X\) be the number of blacks in the sample. (a) Show that \(X\) is approximately a binomial random variable. (b) Check the conditions for using a Normal approximation in this setting. (c) Use the Normal approximation to find the probability that the sample will contain between 165 and 195 blacks. Show your work.

Exercises 47 and 48 refer to the following setting. Two independent random variables \(X\) and \(Y\) have the probability distributions, means, and standard deviations shown. X: 125 P(X): 0.2 0.5 0.3 \(\mu_{X}=2.7, \sigma_{X}=1.55\) Y: 2 4 P(Y): 0.7 0.3 \(\mu_{Y}=2.6, \sigma_{Y}=0.917\) Sum Let the random variable \(T=X+Y\) (a) Find all possible values of \(T\) . Compute the probability that \(T\) takes each of these values. Summarize the probability distribution of \(T\) in a table. (b) Show that the mean of \(T\) is equal to \(\mu_{X}+\mu_{Y}\) (c) Confirm that the variance of \(T\) is equal to \(\sigma_{x}^{2}+\) \(\sigma_{\gamma}^{2}\) Show that \(\sigma_{r} \neq \sigma_{X}+\sigma_{Y}\)

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