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Chapter 6: Problem 18

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

Short Answer

Expert verified
The probabilities are as follows: a) There is a 35% chance that a randomly selected customer will choose a grip size of \(4 \frac{1}{2}\) in. b) There's a 65% chance a customer will choose a racket with a grip size not equal to \(4 \frac{1}{2}\) in. c) The probability \(P(B)\) is the chance that a customer will choose an oversize head racket. d) The probability \(P(C)\) is the chance a customer will choose a racket with a grip size of \(4 \frac{1}{2}\) in. or more. Exact numerical values for \(P(B)\) and \(P(C)\) depend on the missing values in the question.

Step by step solution

01

Interpret the Probability of Event A

The question asks to interpret \(P(A)=.35\). This stands for 'The probability that grip size is \(4 \frac{1}{2}\) inches is 0.35'. Interpretation: There is a 35% chance that a randomly selected customer at the store will choose a tennis racket with a grip size of \(4 \frac{1}{2}\) inches.
02

Calculate the Probability of the complement of Event A

The complement of an event A (denoted \(A'\)) consists of all outcomes not in A. So, to calculate the probability that grip size is not \(4 \frac{1}{2}\) inches, we subtract \(P(A)\) from 1: \(P(A') = 1 - P(A) = 1 - 0.35 = 0.65\). So there's a 65% chance a randomly selected customer will choose a racket with a grip size not equal to \(4 \frac{1}{2}\) inches.
03

Determine the Probability of Event B

The problem might give the probabilities for event B. Then, it's just a matter of summing those probabilities. Similar to Step 1, add up the probabilities of simple outcomes that result in an oversize head. Let's represent this probability as \(P(B)\). The interpretation is: there's a \(P(B)\) chance that a randomly selected customer at the store will choose a racket with an oversize head.
04

Calculate the Probability that Grip Size is at least \(4 \frac{1}{2}\) inches

Find all outcomes where the grip size is \(4 \frac{1}{2}\) inches or more and add their probabilities. Let's denote this probability by \(P(C)\). So, \(P(C)\) is the chance a randomly selected customer will choose a racket with a grip size of \(4 \frac{1}{2}\) in. or more.

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Key Concepts

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

Chance Experiment
A chance experiment is a process or action with a variety of possible outcomes, where the result is determined randomly. Think of it like a game where you can't predict the exact ending each time you play. In our exercise, the chance experiment involves selecting a customer at a store and noting their choice in tennis racket head and grip size. Each time you perform this experiment, you might get different combinations, like different head sizes or grip sizes.

The beauty of a chance experiment lies in its unpredictability. This unpredictability is what makes it a "chance" or "random" experiment. It's important to identify all possible outcomes of the experiment before determining probabilities, as each represents a simple event. In the exercise's context, the simple events are different combinations of racket head sizes and grip sizes.
  • Unpredictability is key.
  • All outcomes are predetermined but occur randomly.
  • Outcomes are also termed as simple events.
Through this framework, probability transforms randomness into a quantifiable metric, allowing us to anticipate the likeliness of each outcome.
Event Outcome
Event outcomes are the specific results we observe after conducting a chance experiment. Each outcome has its probability of occurring. In our scenario, the outcomes concern the combination of a racket's grip size and head size. An event can be a single outcome or a combination of outcomes.

For example, if event A represents the grip size being exactly 4 陆 inches, the probability of this event can be calculated by summing the probabilities of all simple events that fall under event A.
  • The probability of an event is always a number between 0 and 1.
  • Outcomes are the basic components of events.
  • Events can sometimes be described as "favorable" outcomes for a given probability question.
Understanding the probability of event outcomes is crucial when predicting how likely an event is to occur. It's the fundamental building block of probability calculations.
Complementary Probability
Complementary probability refers to the probability of the opposite of an event happening. If you know the probability of an event occurring, you can easily find its complementary probability by subtracting the event's probability from 1. This is because the sum of the probabilities of an event and its complement is always equal to 1.

In our exercise, if there's a 35% chance (0.35 probability) that a racket with a grip size of 4 陆 inches will be chosen, then the probability of choosing a racket with a different grip size would be 1 - 0.35 = 0.65 or 65%.
  • The complement of an event is denoted as \( A' \).
  • The formula: \( P(A') = 1 - P(A) \).
  • This concept is crucial for calculations where the probability of the direct event is not readily available.
By understanding complementary probability, you can deduce missing data points and complete the picture of the scenario's probabilities.
Interpretation of Probability
Interpreting the probability of an event is about understanding what the numerical value means in real-world terms. Probability values provide a measure of how likely an event is to occur and are always expressed as numbers between 0 and 1.

For instance, if the probability of selecting a racket with a grip size of 4 陆 inches is 0.35, this translates to a 35% chance. It implies that in repeated trials of the experiment, 35% of the time, the selected racket will have a grip size of 4 陆 inches.
  • Probabilities closer to 1 indicate a higher likelihood.
  • Probabilities near 0 suggest a lesser chance.
  • Understanding probability anchor decisions, forecasts, and strategies in real-life situations.
Proper interpretation of these values gives weight to decision making, allowing prediction over random outcomes and fostering a better comprehension of data-driven scenarios.

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Most popular questions from this chapter

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

The article "Men, Women at Odds on Gun Control" (Cedar Rapids Gazette, September 8,1999 ) included the following statement: ' The survey found that 56 percent of American adults favored stricter gun control laws. Sixtysix percent of women favored the tougher laws, compared with 45 percent of men." These figures are based on a large telephone survey conducted by Associated Press Polls. If an adult is selected at random, are the events selected adult is female and selected adult favors stricter gun control independent or dependent events? Explain.

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four stu- dents \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives 1 's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? As- suming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?

An article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.
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