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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

The probability distribution of \(x\), the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lrrrrr} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & .54 & .16 & .06 & .04 & .20 \end{array} $$ a. Calculate the mean value of \(x\). b. What is the probability that \(x\) exceeds its mean value?

Because \(P(z<.44)=.67,67 \%\) of all \(z\) values are less than \(.44\), and \(.44\) is the 67 th percentile of the standard normal distribution. Determine the value of each of the following percentiles for the standard normal distribution (Hint: If the cumulative area that you must look for does not appear in the \(z\) table, use the closest entry): a. The 91 st percentile (Hint: Look for area \(.9100 .\) ) b. The 77 th percentile c. The 50 th percentile d. The 9 th percentile e. What is the relationship between the 70 th \(z\) percentile and the 30 th \(z\) percentile?

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one could arrive on Wednesday (W), Thursday (T), Friday (F), or Saturday (S). Suppose that the two magazines arrive independently of one another and that for each magazine \(P(\mathrm{~W})=.4, P(\mathrm{~T})=.3\), \(P(\mathrm{~F})=.2\), and \(P(\mathrm{~S})=.1\). Define a random variable \(y\) by \(y=\) the number of days beyond Wednesday that it takes for both magazines to arrive. For example, if the first magazine arrives on Friday and the second magazine arrives on Wednesday, then \(y=2\), whereas \(y=1\) if both magazines arrive on Thursday. Obtain the probability distribution of \(y\). (Hint: Draw a tree diagram with two generations of branches, the first labeled with arrival days for Magazine 1 and the second for Magazine 2.)

The lifetime of a certain brand of battery is normally distributed with a mean value of \(6 \mathrm{hr}\) and a standard deviation of \(0.8 \mathrm{hr}\) when it is used in a particular cassette player. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. a. What is the probability that the player functions for at least \(4 \mathrm{hr}\) ? b. What is the probability that the cassette player works for at most \(7 \mathrm{hr}\) ? c. Find a number \(z^{*}\) such that only \(5 \%\) of all cassette players will function without battery replacement for more \(\operatorname{than} z^{*} \mathrm{hr}\)

A box contains four slips of paper marked \(1,2,3\), and 4\. Two slips are selected without replacement. List the possible values for each of the following random variables: a. \(x=\) sum of the two numbers b. \(y=\) difference between the first and second numbers c. \(z=\) number of slips selected that show an even number d. \(w=\) number of slips selected that show a 4
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