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Chapter 6: Problem 21

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

Short Answer

Expert verified
Without further information on the relationship between these events, it cannot be definitively stated whether \(P(\text{midsize or } 4 \frac{3}{8} \text{ in. grip}) = .45 + .30 = .75\). The question hinges on whether the events are mutually exclusive or not.

Step by step solution

01

Understanding Mutually Exclusive Events

Two events are said to be mutually exclusive if they cannot both occur at the same time. In this case, if 'midsize' and '4 3/8 inch grip' are mutually exclusive, the probability of either event occurring would indeed be the sum of their individual probabilities.
02

Considering Overlapping Events

If the events can occur simultaneously, they are not mutually exclusive and we risk 'double counting' when we simply add the probabilities. In such a case, the probability of either event occurring is equal to the sum of individual probabilities minus the probability of both events occurring together.
03

Conclusion

To determine whether \(P(\text{midsize or } 4 \frac{3}{8} \text{ in. grip}) = .45 + .30 = .75\) is true, we need additional information about whether these events can occur at the same time. Without this information, we cannot definitively say if the provided equation is accurate.

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Key Concepts

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

Mutually Exclusive Events
In probability theory, understanding mutually exclusive events is crucial. These are events that cannot occur at the same time. For example, when flipping a coin, the result can only be heads or tails, not both鈥攖hese outcomes are mutually exclusive.

Mathematically, if we have two mutually exclusive events, A and B, their combined probability is the sum of their individual probabilities: \( P(A \text{ or } B) = P(A) + P(B) \). This is known as the Addition Rule for Mutually Exclusive Events.

In the problem presented, the term 'midsize' and the specification '4 \frac{3}{8} inch grip' would be considered mutually exclusive if a tennis racket cannot be both simultaneously. If that's the case, then logically, the probability that a racket is either midsize or has a 4 \frac{3}{8} inch grip would indeed be the sum of their individual probabilities, 0.45 and 0.30, respectively.
Overlapping Events
Unlike mutually exclusive events, overlapping events can occur at the same time. These might be two characteristics that are not mutually exclusive, like a person who is both a teacher and a parent.

When dealing with overlapping events, the addition of their probabilities must account for their intersection to avoid double-counting. The correct formula in such cases is: \( P(A \text{ or } B) = P(A) + P(B) - P(A \cap B) \), where \( P(A \cap B) \) is the probability of both A and B occurring together.

In the context of our textbook problem, if the 'midsize' tennis rackets can sometimes have a '4 \frac{3}{8} inch grip', then the events overlap. The probability equation in question would not be correct unless we subtract the probability that a racket is both midsize and has a 4 \frac{3}{8} inch grip. Without that piece of information, we cannot ascertain the true probability of getting either a midsize racket or one with a 4 \frac{3}{8} inch grip.
Probability Theory
Probability theory is a mathematics branch that deals with the likelihood of different events occurring. It is foundational for various fields such as statistics, finance, and risk assessment.

The fundamental principle of probability is that the sum of probabilities for all possible outcomes in a space is 1. When it comes to events, probabilities range between 0 and 1, with 0 indicating an impossible event and 1 a certain one.

Understanding events as mutually exclusive or overlapping is essential in applying probability theory correctly. Students must recognize the type of events they are working with in problems, as this determines the mathematical approach to finding probabilities. The application of the correct formula, whether for mutually exclusive events or overlapping ones, leads to accurate solutions and greater comprehension of the complex scenarios that probability theory often deals with.

Essentially, good grasp of probability theory helps students not just in mathematics, but in making informed decisions based on likelihoods in everyday life.

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Most popular questions from this chapter

A single-elimination tournament with four players is to be held. In Game 1 , the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: \(P(\) seed 1 defeats seed 4\()=.8\) \(P(\) seed 1 defeats seed 2\()=.6\) \(P(\) seed 1 defeats seed 3\()=.7\) \(P(\) seed 2 defeats seed 3\()=.6\) \(P(\) seed 2 defeats seed 4\()=.7\) \(P(\) seed 3 defeats seed 4\()=.6\) a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate Game 3 in the tournament? (This will depend on the outcomes of Games 1 and 2.) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

Suppose that a box contains 25 bulbs, of which 20 are good and the other 5 are defective. Consider randoml selecting three bulbs without replacement. Let \(E\) denote the event that the first bulb selected is good, \(F\) be the event that the second bulb is good, and \(G\) represent the event that the third bulb selected is good. a. What is \(P(E)\) ? b. What is \(P(F \mid E)\) ? c. What is \(P(G \mid E \cap F)\) ? d. What is the probability that all three selected bulbs are good?

Let \(F\) denote the event that a randomly selected registered voter in a certain city has signed a petition to recall the mayor. Also, let \(E\) denote the event that a randomly selected registered voter actually votes in the recall election. Describe the event \(E \cap F\) in words. If \(P(F)=.10\) and \(P(E \mid F)=.80\), determine \(P(E \cap F)\).

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is \(.85 .\) This can be expressed as \(P(\) call is for medical assistance \()=.85\). a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another, calculate the probability that two successive calls will both be for medical assistance. d. Still assuming independence, calculate the prohahility that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think that it is reasonable to assume that the requests made in successive calls are independent? Explain.

\(6.4\) A tennis shop sells five different brands of rackets, each of which comes in either a midsize version or an oversize version. Consider the chance experiment in which brand and size are noted for the next racket purchased. One possible outcome is Head midsize, and another is Prince oversize. Possible outcomes correspond to cells in the following table: $$ \begin{array}{|l|l|l|l|l|l|} \hline & \text { Head } & \text { Prince } & \text { Slazenger } & \text { Wimbledon } & \text { Wilson } \\ \hline \text { Midsize } & & & & & \\ \hline \text { Oversize } & & & & & \\ \hline \end{array} $$ a. Let \(A\) denote the event that an oversize racket is purchased. List the outcomes in \(A\). b. Let \(B\) denote the event that the name of the brand purchased begins with a W. List the outcomes in \(B\). c. List the outcomes in the event \(n o t \bar{B}\). d. Head, Prince, and Wilson are U.S. companies. Let \(C\) denote the event that the racket purchased is made by a U.S. company. List the outcomes in the event \(B\) or \(C\). e. List outcomes in \(B\) and \(C\). f. Display the possible outcomes on a tree diagram, with a first-generation branch for each brand.
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