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There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\), and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F\) ? b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F)\) ? A Venn diagram might help.)

Step by step solution

01

Calculate the Probability of stopping at at least one light

The probability of stopping at least one light is the union of the probabilities of stopping at the first and the second light. This can be calculated with the formula \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). So, we have \(P(E \cup F) = 0.4 + 0.3 - 0.15 = 0.55\).

02

Calculate the Probability of not stopping at either light

The probability of not stopping at either light can be obtained by subtracting the probability of stopping at least one light from the total. Since total probability is 1, we have \(1 - P(E \cup F) = 1 - 0.55 = 0.45\).

03

Calculate the Probability of stopping at exactly one light

The probability of stopping at exactly one light can be calculated as the sum of the probabilities of stopping at each light minus twice the probability of stopping at both lights. This gives us \(P(E) + P(F) - 2P(E \cap F) = 0.4 + 0.3 - 2*0.15 = 0.4\).

04

Calculate the Probability of stopping at just the first light

The probability of stopping just at the first light is the total probability of stopping at the first light minus the probability of stopping at both lights. This gives us \(P(E) - P(E \cap F) = 0.4 - 0.15 = 0.25\).

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Key Concepts

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

Probability of Events

Probability theory is a branch of mathematics focused on analyzing random events. The probability of an event is a measure of the likelihood that the event will occur. Probabilities are expressed as numbers between 0 and 1. An event with a probability of 0 will never occur, while an event with a probability of 1 will always occur. For example, when discussing the probability of stopping at a traffic light, we consider various events such as stopping at the first light (Event E) or the second light (Event F). By knowing the probability of these events, we can make predictions about their occurrence.

• The probability of an event (E) is denoted as \(P(E)\).
• The probabilities show how often each event may happen when they have a chance.
• In the exercise, we have \(P(E) = 0.4\) and \(P(F) = 0.3\).

Union and Intersection of Events

Union and intersection are fundamental concepts in probability theory used to describe combinations of events. The union of two events, \(E \cup F\), occurs if at least one of the events happens. This is often visualized as combining both event probabilities but adjusting for any overlap. The formula for their union is \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\).
For our traffic light scenario, calculating \(P(E \cup F)\) tells us the probability of stopping at at least one light.

• Here, \(P(E \cap F)\) is the probability that both events occur simultaneously, stopping at both lights.
• We found \(P(E \cup F) = 0.55\) using the formula.

The intersection of two events, \(E \cap F\), indicates scenarios where both events happen together. For example, both lights stopping the individual. Given \(P(E \cap F) = 0.15\), it shows the shared chance of both events occurring.

Complementary Events

Complementary events are the counterparts of the events being considered. They represent the scenario where the event does not occur. In probability, the sum of the probabilities of an event and its complement is always 1. For instance, if an event has a probability \(P(E)\), its complement, \(P(E^c)\), can be found as \(1 - P(E)\). For our exercise, complementary events help us understand scenarios such as not stopping at either light.

• For not stopping at any light, we calculate \(1 - P(E \cup F)\) = 0.45.
• This probability represents scenarios where neither of the lights stops the individual.

This complements our understanding of the given problem by showing what it means for events not to occur.

Venn Diagram

A Venn diagram is a useful visual tool in probability theory to represent sets and their relationships with each other through overlapping circles. Each circle represents an event, and the overlaps represent the intersections of events. It makes it easier to see how events combine, intersect, or complement each other.

In our exercise, if we draw a Venn diagram:

• Circle E represents stopping at the first light, while Circle F represents stopping at the second light.
• The overlapping area between E and F shows \(P(E \cap F)\), or stopping at both lights.
• The areas outside the overlap represent stopping at either light exclusively.

This visual representation can clarify the solutions for probabilities such as stopping at exactly one light or stopping at a specific light only. By seeing the problem using a Venn diagram, the relationships and formulas we calculate become much clearer.