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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

Consider a chance experiment that consists of selecting a customer at random from all people who purchased a car at a large car dealership during 2016 . a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.

In a particular state, automobiles that are more than 10 years old must pass a vehicle inspection in order to be registered. This state reports the probability that a car more than 10 years old will fail the vehicle inspection is 0.09 . Give a relative frequency interpretation of this probability.

A large cable company reports the following: \(80 \%\) of its customers subscribe to cable TV service \(42 \%\) of its customers subscribe to Internet service \(32 \%\) of its customers subscribe to telephone service \(25 \%\) of its customers subscribe to both cable TV and Internet service \(21 \%\) of its customers subscribe to both cable TV and phone service \(23 \%\) of its customers subscribe to both Internet and phone service \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. In Exercise \(5.53,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. \(P(\) cable TV only) b. \(P\) (Internet \(\mid\) cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over 6 feet in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is greater, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?
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