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Chapter 2: Problem 14

Suppose that \(55 \%\) of all adults regularly consume coffee, \(45 \%\) regularly consume carbonated soda, and \(70 \%\) regularly consume at least one of these two products. a. What is the probability that a randomly selected adult regularly consumes both coffee and soda? b. What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products?

Short Answer

Expert verified
a: 0.30, b: 0.30

Step by step solution

01

Define the Problem Using Given Percentages

We are given that 55% of adults regularly consume coffee, 45% regularly consume soda, and 70% consume at least one of these two beverages. We need to find the probabilities of various scenarios related to these figures.
02

Translate Percentages into Probabilities

Convert the percentages into probabilities: - Probability of consuming coffee, P(C) = 0.55 - Probability of consuming soda, P(S) = 0.45 - Probability of consuming at least one, P(C 鈭� S) = 0.70.
03

Use the Inclusion-Exclusion Principle

To find the probability of consuming both coffee and soda, P(C 鈭� S), use the Inclusion-Exclusion Principle: \[ P(C 鈭� S) = P(C) + P(S) - P(C 鈭� S) \]Substitute the known values:\[ 0.70 = 0.55 + 0.45 - P(C 鈭� S) \]
04

Solve for Probability of Consuming Both

Rearrange the equation from Step 3 to solve for P(C 鈭� S):\[ P(C 鈭� S) = 0.55 + 0.45 - 0.70 = 0.30 \]This is the probability that an adult regularly consumes both coffee and soda.
05

Calculate Probability of Not Consuming at Least One

The probability that an adult does not regularly consume at least one of the beverages is the complement of consuming at least one:\[ P( ext{Neither C nor S}) = 1 - P(C 鈭� S) = 1 - 0.70 = 0.30 \]
06

Verify Calculation Consistency

Ensure the logic follows through each step: P(C 鈭� S) correctly matches the derived formula and the sum of probabilities is valid given any overlaps existing due to consumers who drink both beverages.

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Key Concepts

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

Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a fundamental concept in probability theory. It is particularly useful when determining the probability of the union of two or more events. This principle helps to calculate the probability of the simultaneous occurrence of two events without double-counting the overlap.
When we talk about two events, let's say event A (consuming coffee) and event B (consuming soda), the principle allows us to find the probability of either A or B or both happening by using the formula:
  • \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \)
Here's what each part of the formula represents:
  • \( P(A) \) is the probability of event A occurring alone.
  • \( P(B) \) is the probability of event B occurring alone.
  • \( P(A \cap B) \) is the probability of both events A and B occurring together, their intersection.
  • Subtracting \( P(A \cap B) \) ensures we don't double-count the overlap.
Using this principle, we can determine the probability of an individual consuming both coffee and soda from the given data.
Probability of Events
Probability of events refers to the likelihood of one or more outcomes occurring within a particular scenario. In this context, it relates to determining how often people consume beverages like coffee and soda. We often express probability as a number between 0 and 1. A probability of 0 means the event never occurs, while a probability of 1 means it always occurs.
In the exercise, probabilities are converted from percentages:
  • \( P(C) = 0.55 \), which is the probability that an adult regularly consumes coffee.
  • \( P(S) = 0.45 \), which is the probability that an adult regularly consumes soda.
  • \( P(C \cup S) = 0.70 \), which is the probability that an adult regularly consumes at least one of these beverages.
These probabilities help us in calculating further probabilities using various principles like Inclusion-Exclusion, aiding in solving parts (a) and (b) of the exercise.
Complementary Probability
Complementary probability is the likelihood of the opposite of a given event happening. This concept is fundamental in solving probability problems because for any event A, the probabilities of A and its complement add up to 1.
Mathematically, it's expressed as:
  • \( P(\text{Not } A) = 1 - P(A) \)
In the exercise, this concept is used to find the probability that a randomly selected adult does not regularly consume either coffee or soda. Known as the complement of consuming at least one, it's calculated by:
  • \( P(\text{Neither C nor S}) = 1 - P(C \cup S) = 1 - 0.70 = 0.30 \)
This calculation shows that there is a 30% probability that an adult does not regularly consume any of the mentioned beverages. Understanding complementary probability provides a quick and efficient way to find such oppositional chances in probability problems.

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Most popular questions from this chapter

According to the article "Optimization of Distribution Parameters for Estimating Probability of Crack Detection" (J. of Aircraft, 2009: 2090-2097), the following "Palmberg" equation is commonly used to determine the probability \(P_{d}(c)\) of detecting a crack of size \(c\) in an aircraft structure: $$ P_{d}(c)=\frac{\left(c / c^{*}\right)^{\beta}}{1+\left(c / c^{*}\right)^{\beta}} $$ where \(c^{*}\) is the crack size that corresponds to a \(.5\) detection probability (and thus is an assessment of the quality of the inspection process). a. Verify that \(P_{d}\left(c^{*}\right)=.5\) b. What is \(P_{d}\left(2 c^{*}\right)\) when \(\beta=4\) ? c. Suppose an inspector inspects two different panels, one with a crack size of \(c^{*}\) and the other with a crack size of \(2 c^{*}\). Again assuming \(\beta=4\) and also that the results of the two inspections are independent of one another, what is the probability that exactly one of the two cracks will be detected? d. What happens to \(P_{d}(c)\) as \(\beta \rightarrow \infty\) ?

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