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Chapter 7: Problem 61

Sophie is a dog that loves to play catch. Unfortunately, she isn't very good, and the probability that she catches a ball is only .1. Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

Short Answer

Expert verified
a. \(x\) has a geometric distribution. b. The probability that it will take exactly two tosses for Sophie to catch a ball is 0.09. c. The probability that more than three tosses will be required is 0.729.

Step by step solution

01

Identifying the Distribution

The prompt states that we're looking for the number of tosses required until Sophie catches a ball; we're not concerned with the number of successes within a given set of trials. This scenario describes a geometric distribution and not a binomial distribution. Therefore, \(x\) has a geometric distribution.
02

Calculating the Probability for Exact Two Tosses

The probability that Sophie catches the ball on her second toss is the product of the probability that she misses the ball on her first toss and the probability that she catches it on the second toss. Since these are independent events, we can use the formula for the geometric probability mass function, \(P(X = x) = (1 - p)^{(x - 1)} * p\), which gives \(P(X = 2) = (1 - 0.1)^{(2 - 1)} * 0.1 = 0.09\).
03

Calculating the Probability for More Than Three Tosses

The problem's asking for the probability that more than three tosses will be required. In other words, at least four tosses. We'll find the probability of the first three tosses, and then subtract that from 1 (representing the total probability). Thus, \(P(X > 3) = 1 - [P(X = 1) + P(X = 2) + P(X = 3)]\). By putting the probabilities for X=1, 2 and 3 in the formula, \(P(X > 3) = 1 - [(0.1) + (0.9*0.1) + (0.9^2*0.1)] = 0.729\).

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Key Concepts

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

Probability
Probability is a measure that indicates how likely an event is to occur. It ranges from 0 to 1, where 0 means the event is impossible, and 1 means the event is certain. In Sophie's case, the probability is used to determine whether she will catch the ball when tossed to her.
To compute specific scenarios, we use simple probabilities. Mathematically, this is often expressed as:
  • Probability of an event happening: \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).
  • Complimentary probability: \( P(A') = 1 - P(A) \), where \( P(A') \) is the probability of the event not occurring.
In our problem, if the probability of Sophie catching the ball is 0.1 on a given toss, then the probability of not catching it is \( 1 - 0.1 = 0.9 \).
By understanding probabilities, we can predict how many attempts might be required for Sophie to finally catch the ball.
Independent Events
Independent events are those not affected by previous outcomes. For instance, Sophie's chances of catching the ball in a new toss are unaffected by previous tosses. This concept is crucial in understanding how each throw remains random, and past failures or successes do not influence future attempts.
  • The probability of independent events occurring in sequence is the product of their individual probabilities. For example, the probability of Sophie not catching the ball twice in a row is \( 0.9 \times 0.9 = 0.81 \).
  • Since each toss is independent, the formula \( P(X = x) = (1 - p)^{(x - 1)} \times p \) can be used for geometric distributions.
Understanding the independence of these events helps in using appropriate statistical formulas to predict Sophie's performance accurately.
Statistics Concepts
Statistics concepts are essential tools in analyzing and interpreting data to make informed decisions. In the context of Sophie catching the ball, you will encounter concepts like expected value, variance, and data prediction. These statistical ideas provide insights into average performance and the consistency surrounding it.
  • Expected Value (Mean): This tells us the long-term average result we can expect if the process is repeated multiple times. For a geometric distribution, the expected value or mean is given by \( \frac{1}{p} \). For example, in this exercise, it is \( \frac{1}{0.1} = 10 \), meaning it would on average take 10 tosses for Sophie to catch the ball once.
  • Variance helps measure how spread out the toss results are around the expected value.
Grasping these statistical concepts allows students to predict outcomes more effectively and understand the variability in Sophie's attempts.
Probability Distributions
A probability distribution tells us the likelihood of all possible outcomes of a random variable. In Sophie's scenario, we are dealing with a geometric distribution, which is perfect for modeling situations where you need to find the 'first success' after a series of independent failures.
  • Geometric Distribution: Here, the variable represents the number of tosses required for Sophie to catch the ball. It focuses on the probability of the first success on the x-th trial.
  • Formula: The geometric distribution formula \( P(X = x) = (1 - p)^{(x - 1)} \times p \) expresses the probability of \( X \) trials needed for the first success. In simple words, it shows the chance that Sophie catches the ball on the x-th attempt.
Understanding probability distributions like the geometric distribution helps us predict when Sophie might catch the ball successfully, allowing for strategic game planning and analysis.

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Most popular questions from this chapter

Consider babies born in the "normal" range of \(37-\) 43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth wcight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal?" The American Statistician [1999]: \(298-302\) ). a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

Suppose that the distribution of net typing rate in words per minute (wpm) for experienced typists can be approximated by a normal curve with mean \(60 \mathrm{wpm}\) and standard deviation 15 wpm ("Effects of Age and Skill in Typing", Journal of Experimental Psychology [1984]: \(345-371\) ). a. What is the probability that a randomly selected typist's net rate is at most 60 wpm? less than 60 wpm? b. What is the probability that a randomly selected typist's net rate is between 45 and 90 wpm? c. Would you be surprised to find a typist in this population whose net rate exceeded 105 wpm? (Note: The largest net rate in a sample described in the paper is 104 wpm.) d. Suppose that two typists are independently selected. What is the probability that both their typing rates exceed 75 wpm? e. Suppose that special training is to be made available to the slowest \(20 \%\) of the typists. What typing speeds would qualify individuals for this training?

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the variable \(x\) as the number of people who actually show up for a sold-out flight. From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{lrrrrrrrr} x & 95 & 96 & 97 & 98 & 99 & 100 & 101 & 102 \\ p(x) & .05 & .10 & .12 & .14 & .24 & .17 & .06 & .04 \\ x & 103 & 104 & 105 & 106 & 107 & 108 & 109 & 110 \\ p(x) & .03 & .02 & .01 & .005 & .005 & .005 & .0037 & .0013 \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

The Los Angeles Times (December 13,1992 ) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

Suppose that \(25 \%\) of the fire alarms in a large city are false alarms. Let \(x\) denote the number of false alarms in a random sample of 100 alarms. Give approximations to the following probabilities: a. \(P(20 \leq x \leq 30)\) b. \(P(20
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