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Chapter 4: Problem 28

Playing darts. Calculate the following probabilities and indicate which probability distribution model is appropriate in each case. A very good darts player can hit the bull's eye (red circle in the center of the dart board) \(65 \%\) of the time. What is the probability that he (a) hits the bullseye for the \(10^{\text {th }}\) time on the \(15^{\text {th }}\) try? (b) hits the bullseye 10 times in 15 tries? (c) hits the first bullseye on the third try?

Short Answer

Expert verified
(a) 0.1225, (b) 0.2816, (c) 0.0796

Step by step solution

01

Identify Probability Distribution Model

In this problem, for parts (a) and (b), we are dealing with a binomial distribution since there's a fixed number of trials (15), two possible outcomes in each trial (hit or miss), the probability of success is constant, and the trials are independent. For part (c), the scenario describes a geometric distribution since it is about determining the number of trials until the first success occurs.
02

Solve for Part (a)

To find the probability of hitting the bullseye for the 10th time on the 15th try, we use the negative binomial distribution formula:\[P(X = k) = \binom{n-1}{k-1} p^k (1-p)^{n-k}\]where \( n = 15 \) is the number of trials, \( k = 10 \) is the required number of successes, and \( p = 0.65 \) is the probability of success.Thus:\[P(X = 10) = \binom{14}{9} (0.65)^{10} (0.35)^{5}\]
03

Solve for Part (b)

To find the probability of hitting the bullseye 10 times in 15 tries, we use the binomial distribution formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( n = 15 \), \( k = 10 \), and \( p = 0.65 \).Thus:\[P(X = 10) = \binom{15}{10} (0.65)^{10} (0.35)^{5}\]
04

Solve for Part (c)

To find the probability of hitting the first bullseye on the third try using a geometric distribution, we use:\[P(X = k) = (1-p)^{k-1} p\]where \( k = 3 \) and \( p = 0.65 \).Thus:\[P(X = 3) = (0.35)^{2} (0.65)\]
05

Calculate Numerical Values

Calculate each probability:**(a)** \[P(X = 10) = \binom{14}{9} (0.65)^{10} (0.35)^{5} \approx 0.1225\]**(b)**\[P(X = 10) = \binom{15}{10} (0.65)^{10} (0.35)^{5} \approx 0.2816\]**(c)**\[P(X = 3) = (0.35)^{2} (0.65) \approx 0.0796\]

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Key Concepts

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

Binomial Distribution
The binomial distribution is a fundamental concept in probability and statistics. It is used to model scenarios where there are a fixed number of independent trials, each with two possible outcomes: success or failure. In situations like the dart player's case, where you're interested in the number of successes (hitting the bullseye) out of a set number of trials (15 tries), the binomial distribution is appropriate. The probability of success is constant for each trial.
  • Formula: The probability mass function (PMF) for a binomial distribution is given by \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where:
    • \( n \) is the number of trials
    • \( k \) is the number of successes
    • \( p \) is the probability of success on an individual trial
  • Example: In the problem, for part (b), we use the binomial distribution to find the probability of hitting the bullseye 10 times in 15 tries.
With this distribution, using the given probability of success \( p = 0.65 \), we can accurately quantify how likely certain numbers of successes are to occur.
Geometric Distribution
The geometric distribution is another probability distribution that is useful in scenarios where you are interested in how many trials it will take to achieve the first success. This distribution has a memoryless property, meaning each trial is independent, and the probability of success remains constant.
  • Formula: For a geometric distribution, the probability of achieving the first success on the \( k^{\text{th}} \) trial can be calculated with\[P(X = k) = (1-p)^{k-1} p\]where:
    • \( k \) is the trial number of the first success
    • \( p \) is the probability of success in each trial
  • Example: In part (c), the focus is on when the first bullseye is hit. We want to find the probability that the player's first success occurs on the third try.
This type of distribution is perfect in instances where intercepting a single event (such as hitting the bullseye for the first time) is crucial. The goal is to understand how the dart player performs in a succession of attempts until they hit their first bullseye.
Probability Calculation
Probability calculation is a cornerstone of statistics and involves determining the likelihood of a particular outcome occurring. The use formulas to ascertain probabilities provided numerical values and insight to predict future events.
  • Understanding Probabilities: Calculating probabilities using distributions such as binomial and geometric requires careful substitution of known values like number of trials, expected successes, and probability of success. The calculations show how likely specific outcomes are.
  • In Practice: In the dart player example, after substituting the values into the respective formulas:
    • For part (a), the result using the negative binomial formula illustrates the probability of the dart player scoring his 10th bullseye on the 15th try.
    • For part (b), using the binomial formula, the result portrays the player hitting exactly 10 bullseyes in 15 tries.
    • For part (c), the geometric formula presents the probability that the player hits the bullseye for the first time on the third try.
    These calculated probabilities are precise and reveal how different distributions fit different problem contexts.
Negative Binomial Distribution
The negative binomial distribution emerges from scenarios where the number of trials required to achieve a fixed number of successes is of interest. Contrary to the binomial distribution, it is not restricted to a fixed number of trials but rather a fixed number of successes.
  • Formula: To calculate probabilities, the formula used is:\[P(X = k) = \binom{n-1}{k-1} p^k (1-p)^{n-k} \]where:
    • \( n \) is the total number of trials
    • \( k \) is the number of successes
    • \( p \) is the probability of success
  • Example Application: In part (a) of the problem, the negative binomial distribution helps find the probability that the player will hit the bullseye for the 10th time on the 15th try. This is particularly useful where the completion of successes spans over several attempts with constant success probability.
The negative binomial distribution effectively models real-world patterns where efforts continue until the desired number of outcomes is reached. This distribution helps articulate scenarios such as part (a) effectively.

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Most popular questions from this chapter

Arachnophobia. A Gallup Poll found that \(7 \%\) of teenagers (ages 13 to 17 ) suffer from arachnophobia and are extremely afraid of spiders. At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. \({ }^{33}\) (a) Calculate the probability that at least one of them suffers from arachnophobia. (b) Calculate the probability that exactly 2 of them suffer from arachnophobia. (c) Calculate the probability that at most 1 of them suffers from arachnophobia. (d) If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, does it seem reasonable for him to randomly assign teenagers to tents?

Triathlon times, Part II. In Exercise 4.4 we saw two distributions for triathlon times: \(N(\mu=4313, \sigma=\) 583 ) for Men, Ages 30 - 34 and \(N(\mu=5261, \sigma=807)\) for the Women, Ages \(25-29\) group. Times are listed in seconds. Use this information to compute each of the following: (a) The cutoff time for the fastest \(5 \%\) of athletes in the men's group, i.e. those who took the shortest \(5 \%\) of time to finish. (b) The cutoff time for the slowest \(10 \%\) of athletes in the women's group.

Stenographer's typos. A very skilled court stenographer makes one typographical error (typo) per hour on average. (a) What probability distribution is most appropriate for calculating the probability of a given number of typos this stenographer makes in an hour? (b) What are the mean and the standard deviation of the number of typos this stenographer makes? (c) Would it be considered unusual if this stenographer made 4 typos in a given hour? (d) Calculate the probability that this stenographer makes at most 2 typos in a given hour.

GRE scores, Part II. In Exercise 4.3 we saw two distributions for GRE scores: \(N(\mu=151, \sigma=7)\) for the verbal part of the exam and \(N(\mu=153, \sigma=7.67)\) for the quantitative part. Use this information to compute each of the following: (a) The score of a student who scored in the \(80^{\text {th }}\) percentile on the Quantitative Reasoning section. (b) The score of a student who scored worse than \(70 \%\) of the test takers in the Verbal Reasoning section.

LA weather, Part I. The average daily high temperature in June in LA is \(77^{\circ} \mathrm{F}\) with a standard deviation of \(5^{\circ} \mathrm{F}\). Suppose that the temperatures in June closely follow a normal distribution. (a) What is the probability of observing an \(83^{\circ} \mathrm{F}\) temperature or higher in LA during a randomly chosen day in June? (b) How cool are the coldest \(10 \%\) of the days (days with lowest average high temperature) during June in \(\mathrm{LA} ?\)
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