/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 There are two traffic lights on ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.)

Short Answer

Expert verified
The probabilities are: 0.55 for stopping at at least one light, 0.45 for not stopping at either light, 0.4 for stopping at exactly one light, and 0.25 for stopping only at the first light.

Step by step solution

01

Probability of Stopping at At Least One Light

To determine the probability that the individual must stop at at least one light, which is denoted as \(E \cup F\), we use the sum rule for probabilities which states that \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\). In substitution, we get \(P(E \cup F) = 0.4 + 0.3 - 0.15 = 0.55.\)
02

Probability of Not Stopping at either Light

To find the probability that the individual doesn't have to stop at either light, we need to find the complement of the event calculated in Step 1. This is given by \(P(\sim(E \cup F)) = 1 - P(E \cup F)\). Substituting the calculated value, we get \(P(\sim(E \cup F)) = 1 - 0.55 = 0.45.\)
03

Probability of Stopping at Exactly One Light

The probability that the individual must stop at exactly one of the two lights is given by the sum of probability of stopping at light one only and light two only. This is mathematically expressed as \(P(E \sim F) + P(\sim E F) = [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]\). Substituting in the given values, we get \(P(E \sim F) + P(\sim E F) = [0.4 - 0.15] + [0.3 - 0.15] = 0.4\).
04

Probability of Stopping Just at the First Light

The probability that the individual stops just at the first light is given by the probability of event \(E\) minus the joint probability of both events \(E\) and \(F\) occurring. This can be mathematically presented as \(P(E \cap \sim F) = P(E) - P(E \cap F)\). Substituting the given values, we have \(P(E \cap \sim F) = 0.4 - 0.15 = 0.25\).

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Key Concepts

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

Conditional Probability
Conditional probability is an important aspect of probability theory that deals with the probability of an event occurring given that another event has already occurred. In the context of the road lights problem, we might consider the probability of needing to stop at the second light given that the individual has already stopped at the first light.

Here, conditional probability is useful when you're focused on how probability changes based on new information. The formula for conditional probability of event B given event A is \[P(B|A) = \frac{P(A \cap B)}{P(A)}.\]Using this concept, one could explore different angles on the problem, such as, "What if the individual knew they stopped at the first light?" It's a useful tool for understanding scenarios where events influence each other.
  • Provides insight into how events relate
  • Useful in decision-making processes
Understanding conditional probability helps us compute the chances in more complex, interrelated situations by considering how one event influences another.
Compound Events
A compound event refers to any event that involves two or more individual events. In probability theory, we use compound events to determine the likelihood of two or more events occurring at the same time or in succession.

In our example with traffic lights, compound event encoding comes into play when we're interested in the probability of stopping at two lights versus one, as well as whether the lights will impact each other. For events E and F, the formula for the compound event that involves both occurrences is given by:\[P(E \cap F) = P(E) \times P(F|E),\]where \(P(F|E)\) is the conditional probability of F occurring given E has occurred. Alternatively, for independent events, it simplifies to:\[P(E \cap F) = P(E) \times P(F).\]
  • Allows for more complex probability calculations
  • Helps to evaluate scenarios with multiple dependent events
Mastering compound events is key to analyzing more advanced probability problems, where multiple variables must be considered at once.
Complementary Events
The notion of complementary events is pivotal when understanding the likelihood of the non-occurrence of an event. In probability, the complement of an event A is denoted as \(\sim A\) and represents all outcomes that are not A.

Simply put, if an event occurs with probability \( P(A) \), then the probability of the complement event (\(\sim A\)) is \[P(\sim A) = 1 - P(A).\]In the example concerning traffic lights, to find the probability that the individual doesn't stop at any of the lights, we use the complement of the union of events E and F ( \( E \cup F \)). This approach helps in evaluating scenarios where we are interested in the exclusion of specific events.
  • Essential for calculating probabilities of non-events
  • Complements other probability rules
Recognizing complementary events broadens our perspective, allowing us to calculate probabilities in a more holistic manner.

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Most popular questions from this chapter

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

6.3 Consider the chance experiment in which the type of transmission- automatic (A) or manual (M) - is recorded for each of the next two cars purchased from a certain dealer. a. What is the set of all possible outcomes (the sample space)? b. Display the possible outcomes in a tree diagram. c. List the outcomes in each of the following events. Which of these events are simple events? i. \(B\) the event that at least one car has an automatic transmission ii. \(C\) the event that exactly one car has an automatic transmission iii. \(D\) the event that neither car has an automatic transmission d. What outcomes are in the event \(B\) and \(C\) ? In the event \(B\) or \(C\) ?

6.12 Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C}\). How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

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