/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 The following table gives a two-... [FREE SOLUTION] | 91Ó°ÊÓ

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The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduated. $$\begin{array}{lcc} \hline & \text { Graduated } & \text { Did Not Graduate } \\ \hline \text { Male } & 126 & 55 \\ \text { Female } & 133 & 32 \\ \hline \end{array}$$ If one of these players is selected at random, find the following probabilities. a. \(P(\) female or did not graduate) b. \(P(\) graduated or male \()\)

Short Answer

Expert verified
a. The probability that a player is female or did not graduate is 0.6358. b. The probability that a player is male or graduated is 0.9075.

Step by step solution

01

Understand the data from the table

From the table, we understand that there are a total of \(126 + 55 + 133 + 32 = 346\) students. The number of females is \(133 + 32 = 165\), and the number of students who did not graduate is \(55 + 32 = 87\). The number of males is \(126 + 55 = 181\), and the number of students who graduated is \(126 + 133 = 259\).
02

Find the probability of \(P(\) female or did not graduate)

The 'or' in probability means 'either or both'. Hence, we add the number of females and the number of students who did not graduate, but this would count twice the female students who did not graduate. So we subtract that number. Hence, \(P(\) female or did not graduate) is \((165 + 87 - 32) / 346 = 220 / 346 = 0.6358\).
03

Find the probability of \(P(\) graduated or male \()\)

Applying the same logic, we add the number of students who graduated and the number of males, but then subtract the male students who graduated, to avoid counting them twice. Hence, \(P(\) graduated or male \()\) is \((259 + 181 - 126) / 346 = 314 / 346 = 0.9075\).

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Key Concepts

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

Two-Way Classification
Two-way classification is a useful method for organizing data into a two-dimensional table, like the one used in this basketball player exercise. This approach categorizes each item according to two attributes—in this case, gender and graduation status. By employing two criteria, it allows for a concise overview of how data is distributed across different groups.
This type of classification helps break down complex information into digestible segments. For instance, in our example, the students are divided into four distinct groups: male and graduated, male but did not graduate, female and graduated, and female but did not graduate. Such tables often use the "row versus column" approach to make data interpretation easier, making it apparent how combinations of two categories manifest among the given population.
Two-way classification tables are crucial in probability because they help us visually and numerically identify which groups overlap or remain exclusive, which further enhances the understanding of the overall dataset.
Conditional Probability
Conditional probability is a measure of the probability of an event occurring under the condition that another event has already happened. In simpler terms, it's concerned with the likelihood of an event given the occurrence of some other related event.
Let's explore with an example: suppose you want to calculate the probability that a player graduated, given that they are male. This calculation looks at the subset of the entire population, in this case, the males, and then determines the fraction of that subset that meets the condition of having graduated.
To compute this, you'd take the ratio of male graduates to all male players, mathematically represented as:\[ P(\text{Graduated} | \text{Male}) = \frac{\text{Male Graduates}}{\text{Total Males}} \]This concept is especially powerful because it helps predict outcomes based on known information, which can refine your understanding of various scenarios.
Combination Rule
The combination rule is commonly linked with problems where events can occur in a variety of ways, and it helps calculate probabilities where multiple possibilities exist. It plays a pivotal role when evaluating probabilities that involve the word "or," signifying that more than one outcome might lead to success.
For example, asking about the probability that a player is female or did not graduate involves overlapping categories—female players and players who did not graduate. While calculating, if we just add the total number of females and those who didn't graduate, we'd be double-counting females who didn't graduate.
To resolve this problem, the combination rule suggests subtracting the intersected count (females who didn't graduate) from the sum to avoid double-counting, as represented by: - Add up numbers in separate groups. - Subtract the overlap (individuals in both groups). This approach ensures precise calculations for probabilities involving "or" conditions.
Independent Events
Independent events in probability are two or more events that do not affect each other's outcomes. This means the occurrence of one event does not influence the probability of the other event taking place.
A classic example of independent events is a coin toss. Flipping a coin and getting heads has no impact on whether you get heads or tails on the next flip. In the context of the table with basketball players, identifying whether two events, like graduation and gender, are independent involves checking if their combined probability matches the product of their individual probabilities.To check independence:\[ P(A \text{ and } B) = P(A) \times P(B) \]If this condition holds, the events are independent.Understanding whether events are independent or dependent can influence decision-making processes and predictions, making it a fundamental concept in statistics.

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Most popular questions from this chapter

In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group (in years). $$\begin{array}{l|lccc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over } &\text { (C) } & 61 & 343 \\ \hline \end{array}$$ a. Suppose one person is randomly selected from these senior citizens. Find the following probabilities. i. \(P(\) have been victimized and \(\mathrm{C}\) ) ii. \(P(\) have never been victimized and \(\mathrm{A}\) ) b. Find \(P(B\) and \(C\) ). Is this probability zero? Explain why or why not.

What is meant by the joint probability of two or more events? Give one example.

Many states have a lottery game, usually called a Pick-4, in which you pick a four-digit number such as 7359 . During the lottery drawing, there are four bins, each containing balls numbered 0 through 9\. One ball is drawn from each bin to form the four-digit winning number. a. You purchase one ticket with one four-digit number. What is the probability that you will win this lottery game? b. There are many variations of this game. The primary variation allows you to win if the four digits in your number are selected in any order as long as they are the same four digits as obtained by the lottery agency. For example, if you pick four digits making the number 1265, then you will win if \(1265,2615,5216,6521\), and so forth, are drawn. The variations of the lottery game depend on how many unique digits are in your number. Consider the following four different versions of this game. i. All four digits are unique (e.g., 1234 ) ii. Exactly one of the digits appears twice (e.g., 1223 or 9095 ) iii. Two digits each appear twice (e.g., 2121 or 5588 ) iv. One digit appears three times (e.g., 3335 or 2722 ) Find the probability that you will win this lottery in each of these four situations.

Two thousand randomly selected adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses obtained $$\begin{array}{lcc} \hline & \text { Have Shopped } & \text { Have Never Shopped } \\ \hline \text { Male } & 500 & 700 \\ \text { Female } & 300 & 500 \\ \hline \end{array}$$ a. Suppose one adult is selected at random from these 2000 adults. Find the following probabilities. i. \(P(\) has never shopped on the Internet and is a male) ii. \(P(\) has shopped on the Internet \(a n d\) is a female) b. Mention what other joint probabilities you can calculate for this table and then find those. You may draw a tree diagram to find these probabilities.

Two thousand randomly selected adults were asked if they are in favor of or against cloning. The following table gives the responses. $$\begin{array}{lccc} \hline & \text { In Favor } & \text { Against } & \text { No Opinion } \\ \hline \text { Male } & 395 & 405 & 100 \\ \text { Female } & 300 & 680 & 120 \\ \hline \end{array}$$ a. If one person is selected at random from these 2000 adults, find the probability that this person is i. in favor of cloning ii. against cloning iii. in favor of cloning given the person is a female iv. a male given the person has no opinion b. Are the events "male" and "in favor" mutually exclusive? What about the events "in favor" and "against?" Why or why not? c. Are the events "female" and "no opinion" independent? Why or why not?

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