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Chapter 6: Problem 91

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid began smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who bas attempted to quit begins smoking again within 2 weeks independent or dependent events? Justify your answer using the given information.

Short Answer

Expert verified
Since the joint probability of the two events \(0.0678\) is not equal to \(0.0704\) which is the product of their individual probabilities, we can conclude that the two events are not independent, but dependent.

Step by step solution

01

Identify needed probabilities

We first need to identify the probability of each event individually. According to the text, the probability that a smoker uses a nicotine aid, P(A), is \(11.3 \%\) or \(0.113\) in decimal form. The probability that someone starts smoking again, P(B), is a combination of the probabilities with or without a nicotine aid, which is \(0.62 * (1 - 0.113) + 0.60 * 0.113 = 0.6233\).
02

Calculate the joint probability

The joint probability of a smoker who uses a nicotine patch and starts smoking again is given by \(P(AB) = P(A) * P(B | A) = 0.113 * 0.60 = 0.0678\).
03

Check for Independence

In the final step, we test whether P(AB) is equal to P(A)*P(B). If they are equal, then events A and B are independent; otherwise, they are dependent. Thus, we compare 0.0678 (the joint probability from Step 2) with 0.113 * 0.6233 = 0.0704.

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Key Concepts

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

Independent Events
In probability theory, independent events are events whose occurrences do not affect each other's outcomes.
Simply put, if knowing that one event has occurred does not change the probability of another event occurring, the events are independent.
To determine if two events, say event A and event B, are independent, we use the principle:
  • If events A and B are independent, then the probability of both A and B occurring (joint probability) is equal to the product of their individual probabilities.
  • Mathematically, this is expressed as: \( P(A \text{ and } B) = P(A) \times P(B) \)
When checking for independence, compare the product of the individual probabilities with the given joint probability.
If they are equal, the events are indeed independent.
Dependent Events
Dependent events in probability theory are events where the occurrence of one event affects the probability of the other event.
This dependency means the probability of one event occurring changes when we know the other event has occurred.
For instance, in the context of trying to quit smoking, the probability of starting to smoke again may change depending upon whether a nicotine aid was used.
  • To establish dependency, one needs to check if \( P(A \text{ and } B) eq P(A) \times P(B) \)
  • Here, \( P(A \text{ given } B) eq P(A) \) can further prove dependency.
In practice, you often have to calculate the probabilities of either scenario. If using a nicotine aid changes the likelihood of starting smoking again, then the events are dependent.
Joint Probability
Joint probability refers to the likelihood of two events occurring at the same time. In simple terms, it answers the question: what is the probability that both A and B happen?
This concept is key when analyzing multiple events that can occur simultaneously.
  • Mathematically, joint probability is denoted by \( P(A \text{ and } B) \)
  • If events are independent, then \( P(A \text{ and } B) = P(A) \times P(B) \)
  • For dependent events, the joint probability is calculated using conditional probability: \( P(A \text{ and } B) = P(A) \times P(B | A) \)
In our smoking scenario, if we select a smoker randomly, the joint probability reflects the probability of that smoker both using a nicotine aid and then relapsing back to smoking within two weeks.

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Most popular questions from this chapter

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.

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?)

The events \(E\) and \(T_{j}\) are defined as \(E=\) the event that someone who is out of work and actively looking for work will find a job within the next month and \(T_{i}=\) the event that someone who is currently out of work has been out of work for \(i\) months. For example, \(T_{2}\) is the event that someone who is out of work has been out of work for 2 months. The following conditional probabilities are approximate and were read from a graph in the paper "The Probability of Finding a Job" (American Economic Review: Papers \& Proceedings [2008]: \(268-273\) ) $$ \begin{array}{ll} P\left(E \mid T_{1}\right)=.30 & P\left(E \mid T_{2}\right)=.24 \\ P\left(E \mid T_{3}\right)=.22 & P\left(E \mid T_{4}\right)=.21 \\ P\left(E \mid T_{5}\right)=.20 & P\left(E \mid T_{6}\right)=.19 \\ P\left(E \mid T_{7}\right)=.19 & P\left(E \mid T_{8}\right)=.18 \\ P\left(E \mid T_{9}\right)=.18 & P\left(E \mid T_{10}\right)=.18 \\ P\left(E \mid T_{11}\right)=.18 & P\left(E \mid T_{12}\right)=.18 \end{array} $$ a. Interpret the following two probabilities: i. \(\quad P\left(E \mid T_{1}\right)=.30\) ii. \(\quad P\left(E \mid T_{6}\right)=.19\) b. Construct a graph of \(P\left(E \mid T_{i}\right)\) versus \(i\). That is, plot \(P\left(E \mid T_{i}\right)\) on the \(y\) -axis and \(i=1,2, \ldots, 12\) on the \(x\) -axis. c. Write a few sentences about how the probability of finding a job in the next month changes as a function of length of unemployment.

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.

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\), and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F\) ? b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F)\) ? A Venn diagram might help.)
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