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Chapter 6: Problem 63

The accompanying probabilities are from the report "Estimated Probability of Competing in Athletics Beyond the High School Interscholastic Level" (www.ncaa.org). The probability that a randomly selected high school basketball player plays NCAA basketball as a freshman in college is \(.0303\); the probability that someone who plays NCAA basketball as a freshman will be playing NCAA basketball in his senior year is .7776; and the probability that a college senior NCAA basketball player will play professionally after college is .0102. Suppose that a high school senior basketball player is chosen at random. Define the events \(F, S\), and \(D\) as \(F=\) the event that the player plays as a college freshman \(S=\) the event that the player also plays as a senior \(D=\) the event that the player plays professionally after college a. Based on the information given, what are the values of \(P(F), P(S \mid F)\), and \(P(D \mid S \cap F)\) ? b. What is the probability that a senior high school basketball player plays college basketball as a freshman and as a senior and then plays professionally? (Hint: This is \(P(F \cap S \cap D) .)\)

Short Answer

Expert verified
The values of \(P(F)\), \(P(S | F)\), and \(P(D | S \cap F)\) are respectively 0.0303, 0.7776, and 0.0102. The probability that a senior high school basketball player plays college basketball as a freshman and as a senior and then plays professionally is approximately 0.00024355776.

Step by step solution

01

Identify given probabilities

The problem states that the probability that a high school basketball player plays NCAA basketball as a freshman in college is 0.0303 (this is \(P(F)\)). The probability that someone who plays NCAA basketball as a freshman plays as a senior is 0.7776 (this is \(P(S | F)\)). Finally, the probability that a college senior NCAA basketball player plays professionally after college is 0.0102 (this is \(P(D | S \cap F)\)).
02

Solve for \(P(F \cap S \cap D)\)

Now, we are required to find the probability that a senior high school basketball player plays basketball as a college freshman, as a senior, and then plays professionally. This can be represented mathematically as \(P(F \cap S \cap D)\). Because these events are sequential (one follows the other), we can multiply the individual probabilities together to obtain the final answer. Therefore, \(P(F \cap S \cap D) = P(F) \cdot P(S | F) \cdot P(D | S \cap F) = 0.0303 \cdot 0.7776 \cdot 0.0102\).
03

Calculate the final probability

Finally, we multiply the three probabilities together to obtain a numerical value: \(0.0303 \cdot 0.7776 \cdot 0.0102 = 0.00024355776\). It's useful to remember that the resulting probability might be a very small number because we are dealing with events that are not very likely. We will round up this final result to 10 decimal places for simplicity.

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Key Concepts

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

Conditional Probability
When we speak of conditional probability, we refer to the likelihood of an event occurring given that another event has already happened. This facet of probability is vital for understanding outcomes in a given context, such as in collegiate athletics. For example, consider the probability of a high school basketball player being active in NCAA basketball by his senior college year. It's not simply a random event 鈥� it depends on whether the player was first admitted and played as a college freshman. In mathematical terms, this probability is denoted as \( P(S | F) \), reading 'the probability of S given F'. The 'given' part is crucial: it indicates that our calculation is grounded on the prior occurrence of event F, a framing which alters the outcome's likelihood compared to considering the event in isolation.
Joint Probability
Joint probability is a term used to describe the probability of two or more events occurring at the same time. It's represented mathematically as \( P(A \text{ and } B) \) or \( P(A \text{ and } B \text{ and } C) \), denoting the likelihood of events A and B (and C) happening together. For our high school basketball player, we want to understand the joint probability of playing NCAA basketball as both a freshman and senior, and then going pro, which involves a chain of events. Here, the joint probability \( P(F \text{ and } S \text{ and } D) \) signifies the cumulative effect of all the previous events鈥� probabilities intersecting.
Probability Theory in Statistics
Probability theory is the mathematical framework that underpins all of statistical analysis. In the field of statistics, probability is used to infer the likelihood of potential outcomes, allowing for the prediction of future events based on past data. This theory is foundational when assessing the career trajectory of student athletes. By examining probabilities, such as the chance of a high school basketball player eventually turning professional, statisticians can provide insights into the patterns and rates of success at various levels of the sport. It gives a quantified perspective on how each stage of an athlete's career influences the next.
Sequential Events Probability
Sequential events probability looks at the likelihood of a series of events happening in a specific order. Within collegiate athletics, this concept can be particularly enlightening when evaluating career paths. The step by step approach seen in the problem showcases sequential probability: starting from a high school player鈥檚 jump to college level (Event F), remaining till their senior year (Event S), and eventually playing professionally (Event D). Each step depends on the successful occurrence of the previous one. With the exercise鈥檚 hint, we understand that we can determine the probability of a high school player making it pro (Events F, S, and D in sequence) by multiplying respective individual probabilities: \( P(F \text{ and } S \text{ and } D) = P(F) \times P(S | F) \times P(D | S \text{ and } F) \). This reflects the fundamental principle that to 'move forward' in a sequence, the previous 'steps' must have been taken.

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Most popular questions from this chapter

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given- say 99 in \(100-\) are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2 ) are first printings and the other three \((3,4\), and 5\()\) are second printings. \(\mathrm{A}\) student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book 5 ?

A company uses three different assembly lines \(A_{1}, A_{2}\), and \(A_{3}\) -to manufacture a particular component. Of those manufactured by \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components and \(10 \%\) of \(A_{3}\) 's components need rework. Suppose that \(50 \%\) of all components are produced by \(A_{1}\), whereas \(30 \%\) are produced by \(A_{2}\) and \(20 \%\) come from \(A_{3}\). a. Construct a tree diagram with first-generation branches corresponding to the three lines. Leading from each branch, draw one branch for rework (R) and another for no rework (N). Then enter appropriate probabilities on the branches. b. What is the probability that a randomly selected component came from \(A_{1}\) and needed rework? c. What is the probability that a randomly selected component needed rework?

The report "Twitter in Higher Education: Usage Habits and Trends of Today's College Faculty" (Magna Publications, September 2009) describes results of a survey of nearly 2000 college faculty. The report indicates the following: \- \(30.7 \%\) reported that they use Twitter and \(69.3 \%\) said that they did not use Twitter. \- Of those who use Twitter, \(39.9 \%\) said they sometimes use Twitter to communicate with students. \- Of those who use Twitter, \(27.5 \%\) said that they sometimes use Twitter as a learning tool in the classroom. Consider the chance experiment that selects one of the study participants at random and define the following events: \(T=\) event that selected faculty member uses Twitter \(C=\) event that selected faculty member sometimes uses Twitter to communicate with students \(L=\) event that selected faculty member sometimes uses Twitter as a learning tool in the classroom a. Use the given information to determine the following probabilities: i. \(\quad P(T)\) ii. \(P\left(T^{C}\right)\) iii. \(P(C \mid T)\) iv. \(P(L \mid T)\) v. \(P(C \cap T)\) b. Interpret each of the probabilities computed in Part (a). c. What proportion of the faculty surveyed sometimes use Twitter to communicate with students? [Hint: Use the law of total probability to find \(P(C) .]\) d. What proportion of faculty surveyed sometimes use Twitter as a learning tool in the classroom?
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