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A total of 28 percent of American males smoke cigarettes, 7 percent smoke cigars, and 5 percent smoke both cigars and cigarettes. (a) What percentage of males smoke neither cigars nor cigarettes? (b) What percentage smoke cigars but not cigarettes?

Short Answer

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(a) 70% of American males smoke neither cigars nor cigarettes. (b) 2% of American males smoke cigars but not cigarettes.

Step by step solution

01

Define the Sets

Let A be the set of American males who smoke cigarettes. Let B be the set of American males who smoke cigars. We are given: \(P(A) = 0.28\) (28% smoke cigarettes) \(P(B) = 0.07\) (7% smoke cigars) \(P(A \cap B) = 0.05\) (5% smoke both cigars and cigarettes)
02

Find the Percentage of Males who Smoke Neither Cigars nor Cigarettes

To find the percentage of males who smoke neither cigars nor cigarettes, we will use the formula for the probability of the union of sets A and B: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Substitute the given values: \(P(A \cup B) = 0.28 + 0.07 - 0.05 = 0.30\) Now, we know the percentage of males who smoke either cigarettes, cigars, or both is 30%. Therefore, the percentage of males who smoke neither cigars nor cigarettes is: \(P(A' \cap B') = 1 - P(A \cup B)\) Calculate the percentage: \(P(A' \cap B') = 1 - 0.30 = 0.70\) So, 70% of American males smoke neither cigars nor cigarettes. Answer (a): 70%.
03

Find the Percentage of Males who Smoke Cigars but not Cigarettes

To find the percentage of males who smoke cigars but not cigarettes, we need to find the percentage of males in set B but not in set A. Using the formula for set difference, we get: \(P(B \setminus A) = P(B) - P(A \cap B)\) Substitute the given values: \(P(B \setminus A) = 0.07 - 0.05 = 0.02\) So, 2% of American males smoke cigars but not cigarettes. Answer (b): 2%.

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Key Concepts

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

Set Theory
Set theory is the mathematical study of collections of objects, known as sets. Imagine a set as a well-defined group of items, which might include numbers, letters, or even people. Each item is called an element of the set.

For instance, in the exercise about smoking habits, we defined two sets: one for American males who smoke cigarettes, and another for those who smoke cigars. Set theory provides the language and the tools to deal with these collections, allowing us to perform operations such as combining sets or finding common elements. It's like having a toolbox that helps us organize and analyze groups where the properties of the whole set matter more than the individual elements.
Union and Intersection of Sets
When we have two sets, we can find what they share in common and how they can be combined. The union of two sets, denoted by \(A \cup B\), includes every element that is in either set A, set B, or in both. It's like creating a guest list for a party by combining the friends you want to invite and the friends your roommate wants to invite. Some friends might be on both lists, but they only get one invitation.

The intersection of sets, denoted by \(A \cap B\), includes only the elements that are in both sets. Continuing our party analogy, it's like finding out which friends are common between you and your roommate's invite lists芒鈧漷hose are the ones who are definitely showing up! In the smoking habits exercise, we looked at American males who both smoke cigarettes and cigars芒鈧漷his is our intersection set.
Complementary Probability
Complementary probability is a concept in probability theory that deals with finding the likelihood of an event not happening. It's based on the idea that the probabilities of all possible outcomes must add up to 1. If you flip a perfectly fair coin, for example, there's a 50% chance it will land on heads and a 50% chance of tails. The complementary probability of not landing on heads (thus, landing on tails) is also 50%.

In our exercise example, when we calculated that 30% of American males smoke either cigarettes, cigars, or both, the complementary probability of 70% represents those who do not partake in smoking at all. This concept is especially useful when it's easier to consider the opposite of an event rather than the event itself.

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Most popular questions from this chapter

Two dice are thrown \(n\) times in succession. Compute the probability that double 6 appears at least once. How large need \(n\) be to make this probability at least \(\frac{1}{2}\) ?

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