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91Ó°ÊÓ

A survey of people in a given region showed that \(20 \%\) were smokers. The probability of death due to lung cancer, given that a person smoked, was roughly 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is \(.006,\) what is the probability of death due to lung cancer given that a person is a smoker?

Short Answer

Expert verified
Answer: The probability of death due to lung cancer given that a person is a smoker is approximately 0.02725 or 2.725%.

Step by step solution

01

Understand the given information

We are given the following probabilities: 1. Probability that a person is a smoker (\(P(S)\)) = 20% or 0.20 2. Probability of death due to lung cancer in the region (\(P(C)\)) = 0.006 3. The probability of death due to lung cancer given that a person smoked (\(P(C|S)\)) is 10 times the probability of death due to lung cancer, given that a person did not smoke (\(P(C|S^{'})\)), i.e., \(P(C|S) = 10 * P(C|S^{'})\).
02

Apply the conditional probability formula

To compute the probability of death due to lung cancer given that a person is a smoker, we will use the following conditional probability formula: \(P(C|S) = \frac{P(C \cap S)}{P(S)}\) We know \(P(S)\), but we need to find \(P(C \cap S)\).
03

Compute the probability of intersection

We can use the formula \(P(C) = P(C \cap S) + P(C \cap S^{'})\) to compute the intersection probability. We know \(P(C)\), and we can write the unknowns in terms of each other as follows: \(P(C \cap S) = 10 * P(C \cap S^{'})\) \(P(C \cap S) + P(C \cap S^{'}) = 0.006\) Now we can solve for \(P(C \cap S)\).
04

Solve for the intersection probability

To solve for \(P(C \cap S)\), we can use the system of equations from Step 3: 1. \(P(C \cap S) = 10 * P(C \cap S^{'})\) 2. \(P(C \cap S) + P(C \cap S^{'}) = 0.006\) Substitute the first equation into the second equation: \(10 * P(C \cap S^{'}) + P(C \cap S^{'}) = 0.006 \Rightarrow 11 * P(C \cap S^{'}) = 0.006\) Solving for \(P(C \cap S^{'})\): \(P(C \cap S^{'}) = \frac{0.006}{11} \approx 0.000545\) Now we can calculate \(P(C \cap S)\): \(P(C \cap S) = 10 * P(C \cap S^{'}) = 10 * 0.000545 \approx 0.00545\)
05

Calculate the probability of death due to lung cancer given that a person is a smoker

Now we can calculate \(P(C|S)\) using the formula: \(P(C|S) = \frac{P(C \cap S)}{P(S)} = \frac{0.00545}{0.20} \approx 0.02725\) Therefore, the probability of death due to lung cancer given that a person is a smoker is approximately 0.02725 or 2.725%.

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Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a fundamental concept in probability theory and statistics. It allows us to update the probability estimate of an event, based on additional information or evidence. This theorem connects conditional probabilities and provides a way to reverse them.
In the context of the exercise, we wanted to find the probability of death due to lung cancer given that a person is a smoker. While the problem did not directly state the use of Bayes' Theorem, the step-by-step approach overlaps with its conceptual application. By iteratively expressing joint and conditional probabilities, we aimed to find a more accurate probability based on given conditions.
Bayes' Theorem is given by:
\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]
In our problem, "A" could represent the event of a person being a smoker, and "B" represents the event of death due to lung cancer. But, we often rearrange and adapt formulas in practical problems, as seen in the exercise. Understanding this theorem is essential because it forms the backbone of many probability-based analyses and helps contextualize everyday scenarios with structured mathematical reasoning.
Probability Distributions
Probability distributions provide a complete description of the probability of different outcomes in a random process. They serve as a roadmap for understanding how probabilities are distributed among all possible outcomes of an event.
In this exercise, we can think of smokers and non-smokers as categories within a probability distribution describing the likelihood of lung cancer-related deaths. The probabilities provided form part of the description of this distribution.
Suppose you visualize a probability distribution. Smokers might represent a "spike" in this distribution for lung cancer deaths, assuming a much higher likelihood than non-smokers. These spikes can be estimated and calculated using statistical methods, thus showing how smoking status drastically changes the risk distribution for lung cancer.
Understanding probability distributions helps us recognize patterns and relationships, such as how a small percentage of smokers might account for a relatively larger percentage of lung cancer-related deaths compared to non-smokers.
Statistical Analysis
Statistical analysis involves collecting, analyzing, interpreting, presenting, and organizing data. It’s essential in science, business, and social sciences to make informed decisions based on data sets.
In the context of this exercise, statistical analysis enables us to examine the relationship between smoking and lung cancer. By analyzing data, we can quantify the increased risk smokers face and explore the implications of different factors influencing those risks.
Conduct statistical analysis by:
  • Identifying the data variables: Here, the data involves the smoking status and lung cancer occurrence.
  • Applying probability concepts: Use probability formulas, such as the conditional probability and intersection probability shown in the solution.
  • Interpreting results: After calculations, the results provide insights into risk levels and guide health recommendations.

Statistical analysis not only gives us numbers and percentages but also aids in understanding the deeper story behind data. It’s crucial for predicting outcomes and altering policies or behaviors based on comprehensive data interpretations.

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Most popular questions from this chapter

Suppose that at a particular supermarket the probability of waiting 5 minutes or longer for checkout at the cashier's counter is .2. On a given day, a man and his wife decide to shop individually at the market, each checking out at different cashier counters. They both reach cashier counters at the same time. a. What is the probability that the man will wait less than 5 minutes for checkout? b. What is probability that both the man and his wife will be checked out in less than 5 minutes? (Assume that the checkout times for the two are independent events.) c. What is the probability that one or the other or both will wait 5 minutes or longer?

Permutations Evaluate the following permutations. a. \(P_{3}^{5}\) b. \(P_{9}^{10}\) c. \(P_{6}^{6}\) d. \(P_{1}^{20}\)

Two fair dice are tossed. Use the Tossing Dice applet to answer the following questions. a. What is the probability that the sum of the number of dots shown on the upper faces is equal to \(7 ?\) To \(11 ?\) b. What is the probability that you roll "doubles" that is, both dice have the same number on the upper face? c. What is the probability that both dice show an odd number?

The American Journal of Sports Medicine published a study of 810 women collegiate rugby players who have a history of knee injuries. For these athletes, the two common knee injuries investigated were medial cruciate ligament (MCL) sprains and anterior cruciate ligament (ACL) tears. \(^{8}\) For backfield players, it was found that \(39 \%\) had MCL sprains and \(61 \%\) had ACL tears. For forwards, it was found that \(33 \%\) had MCL sprains and \(67 \%\) had ACL tears. Since a rugby team consists of eight forwards and seven backs, you can assume that \(47 \%\) of the players with knee injuries are backs and \(53 \%\) are forwards. a. Find the unconditional probability that a rugby player selected at random from this group of players has experienced an MCL sprain. b. Given that you have selected a player who has an MCL sprain, what is the probability that the player is a forward? c. Given that you have selected a player who has an ACL tear, what is the probability that the player is a back?

A salesperson figures that the probability of her consummating a sale during the first contact with a client is .4 but improves to .55 on the second contact if the client did not buy during the first contact. Suppose this salesperson makes one and only one callback to any client. If she contacts a client, calculate the probabilities for these events: a. The client will buy. b. The client will not buy.

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