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Chapter 5: Problem 57

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

Short Answer

Expert verified
a. The probability that Shelly must stop for at least one light is \(P(E \cup F) = 0.55\). b. The probability that Shelly does not have to stop at either light is \(P((E \cup F)') = 0.45\). c. The probability that Shelly must stop at exactly one of the two lights is \(0.4\). d. The probability that Shelly must stop only at the first light is \(0.25\).

Step by step solution

01

Use the formula for \(P(E \cup F)\)

To calculate the probability of Shelly stopping for at least one light, we need to find the probability of the event \(E \cup F\). We can do this using the formula \(P(E \cup F)= P(E)+P(F)-P(E \cap F)\).
02

Plug in the given probabilities

Using the given probabilities, we can plug them into the formula to get \(P(E \cup F) = 0.4 + 0.3 - 0.15\).
03

Calculate the probability

Now, performing the arithmetic, we get the probability of Shelly stopping for at least one light: \(P(E \cup F) = 0.55\). b. The probability that Shelly does not have to stop at either light.
04

Use the complement rule

To find the probability that Shelly does not have to stop at either light, we need to find the probability of the event \((E \cup F)'\). Since \((E \cup F)'\) is the complement of \(E \cup F\), we know that \(P((E \cup F)') = 1 - P(E \cup F)\).
05

Plug in the probability of \(E \cup F\)

Using the probability of \(E \cup F\) calculated in part a, we get \(P((E \cup F)') = 1 - 0.55\).
06

Calculate the probability

Now, performing the arithmetic, we get the probability of Shelly not stopping at either light: \(P((E \cup F)') = 0.45\). c. The probability that Shelly must stop at exactly one of the two lights.
07

Use the formula for probability of stopping at only one light

To find the probability of stopping at exactly one light, we can use the fact that \(P(E \cup F) - P(E\cap F)\). Since we already found the probability of stopping at least one light, we can just subtract the probability of stopping at both lights.
08

Calculate the probability

Using the probabilities calculated in the previous steps, we get the probability of stopping at exactly one light as: \(0.55 - 0.15 = 0.4\). d. The probability that Shelly must stop only at the first light.
09

Use the formula for probability of stopping at only the first light

To find the probability of stopping only at the first light, we can use the fact that \(P(E) - P(E \cap F)\). Since we have already been given the probability of stopping at the first light and the probability of stopping at both lights, we just subtract the latter from the former.
10

Calculate the probability

Using the given probabilities, we get the probability of stopping only at the first light as: \(0.4 - 0.15 = 0.25\).

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Key Concepts

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

Probability Formulas
Probability formulas are essential mathematical tools that help us determine the likelihood of events occurring. When dealing with multiple events, such as whether Shelly has to stop at traffic lights, these formulas allow us to calculate the probability of various outcomes.

The probability of an event, denoted by \(P(A)\), is a number between 0 and 1 that indicates how likely it is that the event will happen. Values closer to 1 indicate a higher likelihood, while values near 0 suggest a lower likelihood.

For example, to find the probability that Shelly stops at at least one traffic light (either the first or second), we use a formula for the union of two events \(E\) and \(F\), \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). This accounts for the probability of each event happening while subtracting the overlap, or joint probability of both events happening together.
Complement Rule
In probability, the complement rule is a handy concept used to find the probability of the opposite of a given event. The complement of an event \(A\), denoted \(A'\), consists of all possible outcomes in the sample space that are not part of \(A\).

The probability of the complement, \(P(A')\), is calculated as \(1 - P(A)\). This is because all probabilities in a sample space add up to 1.

In Shelly's scenario, the complement rule helps us find the probability that she doesn't have to stop at either of the traffic lights. By knowing the probability of stopping at least once, \(P(E \cup F)\), she can calculate the complement probability, \(P((E \cup F)')\), which is \(1 - P(E \cup F)\). This means we simply subtract the probability of the union of events from 1 to find the complement.
Intersection of Events
The intersection of two events, denoted by \(E \cap F\), is a key concept in probability that refers to the scenario where both events occur simultaneously. Knowing the probability of the intersection helps in determining more complex situations.

In our example, \(P(E \cap F)\) represents the probability that Shelly has to stop at both traffic lights on her way to work. It is given as 0.15.

This probability is critical when computing other probabilities, such as the probability of stopping at exactly one light. We use the intersection to adjust for the overlap portion when calculating the probability of the union of events.
Union of Events
The union of events \(E\) and \(F\), expressed as \(E \cup F\), represents the probability that at least one of the events will occur. This could mean either \(E\) occurs, \(F\) occurs, or both \(E\) and \(F\) occur.

To find the probability of the union, we employ the formula \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). This formula accounts for the addition of the probabilities of the individual events while subtracting the probability that both events occur together to avoid double-counting.

In Shelly鈥檚 commute, finding \(P(E \cup F) = 0.55\) tells us that there's a 55% chance she will be stopped at one or more lights. This method ensures we accurately assess the likelihood of encountering at least one red light during her trip.

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Most popular questions from this chapter

Airline tickets can be purchased online, by telephone, or by using a travel agent. Passengers who have a ticket sometimes don't show up for their flights. Suppose a person who purchased a ticket is selected at random. Consider the following events: \(O=\) event selected person purchased ticket online \(N=\) event selected person did not show up for flight Suppose \(P(O)=0.70, P(N)=0.07,\) and \(P(O \cap N)=0.04\) a. Are the events \(N\) and \(O\) independent? How can you tell? b. Construct a hypothetical 1000 table with columns corresponding to \(N\) and \(\operatorname{not} N\) and rows corresponding to \(O\) and \(\operatorname{not} O\). c. Use the table to find \(P(O \cup N)\). Give a relative frequency interpretation of this probability.

Each time a class meets, the professor selects one student at random to explain the solution to a homework problem. There are 40 students in the class, and no one ever misses class. Luke is one of these students. What is the probability that Luke is selected both of the next two times that the class meets?

A company that offers roadside assistance to drivers reports that the probability that a call for assistance will be to help someone who is locked out of his or her car is \(0.18 .\) Give a relative frequency interpretation of this probability.

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The report "Improving Undergraduate Learning" (Social Science Research Council, 2011) summarizes data from a survey of several thousand college students. These students were thought to be representative of the population of all college students in the United States. When asked about an upcoming semester, \(68 \%\) said they would be taking a class that is reading-intensive (requires more than 40 pages of reading per week). Only \(50 \%\) said they would be taking a class that is writing-intensive (requires more than 20 pages of writing over the course of the semester). The percentage who said that they would be taking both a reading-intensive course and a writing-intensive course was \(42 \%\). a. Use the given information to set up a hypothetical 1000 table. b. Use the table to find the following probabilities: i. the probability that a randomly selected student would be taking at least one reading-intensive or writing-intensive course. ii. the probability that a randomly selected student would be taking a reading-intensive course or a writingintensive course, but not both. iii. the probability that a randomly selected student would be taking neither a reading-intensive nor a writing-intensive course.
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