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91Ó°ÊÓ

According to an Automobile Association of America report, \(9.6 \%\) of Americans traveled by car over the 2011 Memorial Day weekend and \(88.09 \%\) stayed home. What is the probability that a randomly selected American stayed home or traveled by car over the 2011 Memorial Day weekend? Explain why this probability does not equal 1.0.

Short Answer

Expert verified
The probability that a randomly selected American either stayed home or traveled by car over the 2011 Memorial Day weekend is 0.9769 or 97.69%. This probability does not equal 1.0 because the problem does not account for Americans who may have used other modes of transportation during that weekend.

Step by step solution

01

Convert Percentages to Probabilities

Given percentages need to be converted into probabilities. From the problem statement, 9.6% of Americans traveled by car and 88.09% stayed at home. This can be represented as probabilities by dividing the percentages by 100. Therefore, the probability of an American traveling by car is 0.096 and the probability of an American staying home is 0.8809.
02

Calculate Combined Probability

To find the probability that a randomly selected American stayed home or traveled by car, we simply add the two probabilities together. The 'or' in probability refers to the union of events, which is the sum of the probabilities of the individual events. Therefore, the combined probability is 0.096 + 0.8809 = 0.9769.
03

Explain why the Combined Probability Does Not Equal 1.0

Probability values can range from 0 to 1, where 0 means an event is impossible and 1 means it is certain to happen. Normally, if we had considered all possible events, the total probability should be 1. However, in this scenario, the calculated probability does not add up to 1 as the problem doesn't account for other modes of transportation. The remaining 2.31% could refer to other means of travel, like planes or trains.

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Key Concepts

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

Percentages to Probabilities
In probability, we often start with percentages that express how likely an event is.
To convert these percentages into probabilities, we simply divide them by 100.
This places the percentage into a decimal form, suitable for probability calculations.

For example, if you know that 9.6% of people traveled by car, you convert this to a probability by calculating \( \frac{9.6}{100} = 0.096 \).
Similarly, for those who stayed home which is given as 88.09%, the probability would be \( \frac{88.09}{100} = 0.8809 \).

These conversions are essential because probabilities in mathematics are always between 0 and 1.
0 means an event will not happen, and 1 means it will definitely happen.
Therefore, percentages allow for a clear bridge to understanding and utilizing these concepts in practical problem-solving.
Combined Probability
In situations where you need to find the probability of one event happening or another, it's about combining probabilities.
This is often referred to as finding the "union" of two events.
For example, what's the chance of "either traveling by car or staying home"? We find this by adding the individual probabilities.

In our scenario, we calculated:
  • The probability of traveling by car: 0.096
  • The probability of staying at home: 0.8809
Add them together, 0.096 + 0.8809, to get a combined probability of 0.9769.
This calculation represents the likelihood of one event or the other happening, known as their union.
It's useful for understanding how often one of several different outcomes might occur.
Union of Events
Understanding the union of events is key in probability.
When you ask about the union, you're essentially asking what the chance is that at least one of several events will happen.
This is crucial when considering multiple possibilities.

In terms of our example, calculating the union of staying home or traveling by car was straightforward.
However, the total probability of all possible events must equal 1.
In our solution, it didn't, because we didn't account for other modes of travel, like air or train.
This leftover percentage accounts for the rest of the possibilities we didn't include.

Realizing this helps us understand how to include all outcomes and recognize gaps in our initial assumptions.
The union of all possible events indeed sums up to 1, ensuring a complete evaluation of the situation.

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Most popular questions from this chapter

The probability that a student graduating from Suburban State University has student loans to pay off after graduation is .60. The probability that a student graduating from this university has student loans to pay off after graduation and is a male is \(.24\). Find the conditional probability that a randomly selected student from this university is a male given that this student has student loans to pay off after graduation.

Given that \(P(A)=.72\) and \(P(A\) and \(B)=.38\), find \(P(B \mid A)\).

Find the joint probability of \(A\) and \(B\) for the following. a. \(P(B)=.66\) and \(P(A \mid B)=.91\) b. \(P(A)=.12\) and \(P(B \mid A)=.07\)

A consumer agency randomly selected 1700 flights for two major airlines, \(\mathrm{A}\) and \(\mathrm{B}\). The following table gives the two-way classification of these flights based on airline and arrival time. Note that "less than 30 minutes late" includes flights that arrived early or on time. $$ \begin{array}{lccc} \hline & \begin{array}{c} \text { Less Than 30 } \\ \text { Minutes Late } \end{array} & \begin{array}{c} \text { 30 Minutes to } \\ \text { 1 Hour Late } \end{array} & \begin{array}{c} \text { More Than } \\ \text { 1 Hour Late } \end{array} \\ \hline \text { Airline A } & 429 & 390 & 92 \\ \text { Airline B } & 393 & 316 & 80 \\ \hline \end{array} $$ If one flight is selected at random from these 1700 flights, find the following probabilities. a. \(P\) (more than 1 hour late or airline \(\mathrm{A}\) ) b. \(P(\) airline \(B\) or less than 30 minutes late) c. \(P(\) airline A or airline \(\mathrm{B}\) )

Forty-seven employees in an office wear eyeglasses. Thirty-one have single- vision correction, and 16 wear bifocals. If two employees are selected at random from this group, what is the probability that both of them wear bifocals? What is the probability that both have single-vision correction?

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