91Ó°ÊÓ

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 \(P(E \cap F)=0.15 .\) (Hint: See Example 5.5) a. Use the given probability information to set up a "hypothetical \(1000 "\) table with columns corresponding to \(E\) and not \(E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. the probability that Shelly must stop for at least one light (the probability of \(E \cup F)\). ii. the probability that Shelly does not have to stop at either light. iii. the probability that Shelly must stop at exactly one of the two lights. iv. the probability that Shelly must stop only at the first light.

Step by step solution

01

Set up the hypothetical table

Start with a total of 1000. Multiply the total by the appropriate given probabilities to find the respective frequencies. The value in the \(E\) and \(F\) cell is given as 0.15*1000 = 150. The value in the \(E\) and not \(F\) cell is \(P(E) - P(E \cap F)\) = 0.4 - 0.15 = 0.25*1000 = 250. Similarly, the value in the \(F\) and not \(E\) cell is \(P(F) - P(E \cap F)\) = 0.3 - 0.15 = 0.15*1000 = 150. The value in the not \(E\) and not \(F\) cell is Total - ((\(E and F\)) + (\(E\) and not \(F\)) + (not \(E\) and \(F\))) = 1000 - (150+250+150) = 450. Complete the marginal totals for each row and column to yield frequency for \(E\), not \(E\), \(F\), and not \(F\).

02

The probability of \(E \cup F\)

This means that Shelly must stop for at least one light. This is equal to \(P(E) + P(F) - P(E \cap F)\) = 0.4 + 0.3 - 0.15 = 0.55. Dividing the obtained frequency by total (1000), 0.55 is indeed the sum of probabilities of all the cases where Shelly stops at least at one light: \(E and F\), \(E and not F\), and \(F and not E\)

03

The probability that Shelly does not have to stop at either light

This is the value in the not \(E\) and not \(F\) cell. Therefore the probability is 450/1000 = 0.45.

04

The probability that Shelly must stop at exactly one of the two lights

This is the sum of \(E\) and not \(F\), and \(F\) and not \(E\). Therefore, the probability is (250+150)/1000 = 0.4

05

The probability that Shelly must stop only at the first light

This is the \(E\) and not \(F\) event. Therefore the probability is 250/1000 = 0.25

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Key Concepts

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

Probability of Events

Understanding how likely it is for certain events to occur is the essence of probability theory. When we talk about the probability of an event, we are quantifying the chance that the event will happen. For instance, if Shelly must stop at the first traffic light on her route to work, denoted as event E, the probability of this occurring is represented mathematically as P(E).

In the given example exercise, it's stated that P(E) = 0.4, which means there is a 40% chance that Shelly will have to stop at the first light during her journey. This percentage is calculated based on predefined conditions or historical data. Moreover, the exercise guides students to visualize probabilities using a hypothetical table of 1000 instances which can be particularly effective for grasping the concept of probability, making abstract ideas more concrete.

Intersection of Events

When dealing with multiple events, it's often necessary to consider the probability that all the events occur together. The intersection of events, denoted by the symbol ∩, represents the set of outcomes where both events happen at the same time. For Shelly's commute, the event where she must stop at both the first light, E, and the second light, F, is an intersection of events, represented as E ∩ F.

In probability terms, if P(E ∩ F) = 0.15, it indicates a 15% chance of Shelly stopping at both lights. The exercise utilizes the intersection concept to further analyze the combined probability, enhancing the student's comprehension of how events relate to each other in probabilistic terms.

Union of Events

Conversely, we may want to know the probability of at least one of several events occurring. This is where the union of events comes in, indicated by the symbol ∪. The probability of the union of events, E ∪ F, is the chance that either event E occurs, or event F occurs, or they both occur.

For Shelly's route, P(E ∪ F) represents the likelihood of her needing to stop at one or both lights. To calculate this, we add the probabilities of each event and subtract the probability of their intersection: P(E) + P(F) - P(E ∩ F). The exercise demonstrates this calculation and allows students to explore the probability of combined events in a practical context.

Conditional Probability

Conditional probability is concerned with the probability of an event occurring given that another event has already occurred. This can be a bit more complex because the occurrence of the first event affects the probability of the second event. While the original exercise doesn't explicitly ask for conditional probability, understanding this concept is critical when events are dependent on each other.

For example, if we wanted to know the probability that Shelly stops at the second light given she has already stopped at the first, we would calculate the conditional probability as P(F|E). This is determined by the ratio of the probability of both events happening to the probability of the first event, or P(E ∩ F) / P(E). This concept is essential for more complex probability scenarios and deepens a student’s analytical skills in the realm of potential outcomes.