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Chapter 6: Problem 101

Suppose that a box contains 25 light bulbs, of which 20 are good and the other 5 are defective. Consider randomly 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?

Short Answer

Expert verified
a) \(P(E) = 0.8\) \n b) \(P(F | E) = 0.79167\) \n c) \(P(G | E \cap F) = 0.78261\) \n d) The probability that all three selected bulbs are good is \(0.49688\).

Step by step solution

01

Calculate \(P(E)\)

The event \(E\) is that the first bulb selected is good. Since there are 20 good bulbs out of a total of 25, the probability that the first bulb selected is good, \(P(E)\), is simply the ratio of good bulbs to total bulbs, written mathematically as \(P(E) = \frac{20}{25} = 0.8\).
02

Calculate \(P(F | E)\)

The event \(F\) is that the second bulb is good and it is conditional on event \(E\), the first bulb being good. If the first bulb was good, we have 19 good bulbs left of a total of 24 bulbs, thus \(P(F|E) = \frac{19}{24} = 0.79167.\)
03

Calculate \(P(G | E \cap F)\)

The event \(G\) is that the third bulb is good and it is conditional on both the first bulb and the second bulb being good. If the first two bulbs were good, we have 18 good bulbs left out of a total of 23 bulbs, thus \(P(G|E \cap F) = \frac{18}{23} = 0.78261\) .
04

Calculate the probability that all three selected bulbs are good

The probability that all three selected bulbs are good is simply the multiplication of the probabilities of each individual event, since they are dependent events: \(\Pi P (good) = P(E) \times P(F|E) \times P(G|E \cap F) = 0.8 \times 0.79167 \times 0.78261 = 0.49688.\)

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in statistics that deals with the likelihood of an event occurring given that another event has already occurred. Think of it as updating your expectations based on new information.

Regarding the light bulb example, once we know the first bulb (\(E\)) is good, the proportion of good bulbs out of the remaining total changes. The conditional probability of selecting a good second bulb (\(P(F | E)\)) is different from the initial unconditional probability because our sample space has been updated: we're now choosing from 24 bulbs instead of the original 25, with one less good bulb to pick.
Dependent Events
Dependent events in probability are those whose outcomes are linked. The likelihood of a subsequent event changes based on the outcomes of the preceding ones.

In the context of our non-replacement light bulb scenario, each selection affects the next. If the first bulb selected is good (\(E\)), this affects the probability of the subsequent bulb (\(F\)) also being good. The events are dependent because the selection of each bulb alters the composition of what remains in the box, influencing the odds for the next draw.
Probability Theory
Probability theory is the mathematical framework that describes the nature of random events. It helps to predict the likeliness of outcomes in situations where there is some form of uncertainty.

Learning probability theory involves understanding concepts like randomness, events, sample spaces, and probabilities themselves, as seen in our exercise. For each bulb picked, probability theory gives us the tools to determine how likely it is to be good or defective, establishing a systematic way to make informed predictions based on the evidence at hand.
Combinatorics
Combinatorics is the branch of mathematics dealing with counting and arrangements of objects. It's closely related to probability, as it provides methods to enumerate possibilities, which is critical for calculating probabilities in more complex scenarios.

While not directly involved in our step-by-step light bulb example, understanding combinatorics would be essential if, for instance, we wanted to know how many ways we could select three bulbs in total. It enables us to calculate the total number of outcomes and thus the likelihood of any specific outcome occurring.

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Most popular questions from this chapter

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.

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for \(A\), a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3\), \(P(\) B wins game \()=.2\), and \(P(\) draw \()=.5\). Each time a player wins a game, he earns 1 point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that \(A\) wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation to estimate \(\mathrm{P}(\mathrm{A}\) wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

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 students \((1,2,3\), 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 l'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? Assuming 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?

Delayed diagnosis of cancer is a problem because it can delay the start of treatment. The paper "Causes of Physician Delay in the Diagnosis of Breast Cancer" (Archives of Internal Medicine \([2002]: 1343-1348)\) examined possible causes for delayed diagnosis for women with breast cancer. The accompanying table summarizes data on the initial written mammogram report (benign or suspicious) and whether or not diagnosis was delayed for 433 women with breast cancer. $$ \begin{array}{l|cc} & & \text { Diagnosis } \\ & \begin{array}{c} \text { Diagnosis } \\ \text { Delayed } \end{array} & \begin{array}{c} \text { Not } \\ \text { Delayed } \end{array} \\ \hline \begin{array}{l} \text { Mammogram Report Benign } \\ \text { Mammogram Report } \\ \text { Suspicious } \end{array} & 32 & 89 \\ & 8 & 304 \\ \hline \end{array} $$ Consider the following events: \(B=\) the event that the mammogram report says benign \(S=\) event that the mammogram report says suspicious \(D=\) event that diagnosis is delayed a. Assume that these data are representative of the larger group of all women with breast cancer. Use the data in the table to find and interpret the following probabilities: i. \(\quad P(B)\) ii. \(P(S)\) iii. \(P(D \mid B)\) iv. \(P(D \mid S)\) b. Remember that all of the 433 women in this study actually had breast cancer, so benign mammogram reports were, by definition, in error. Write a few sentences explaining whether this type of error in the reading of mammograms is related to delayed diagnosis of breast cancer.

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)
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