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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

An article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a schedule is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only . 6 . 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only .5. 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only \(.7 .\) Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

Of the 10,000 students at a certain university, 7000 have Visa cards, 6000 have MasterCards, and 5000 have both. Suppose that a student is randomly selected. a. What is the probability that the selected student has a Visa card? b. What is the probability that the selected student has both cards? c. Suppose you learn that the selected individual has a Visa card (was one of the 7000 with such a card). Now what is the probability that this student has both cards? d. Are the events has \(a\) Visa card and has a MasterCard independent? Explain. e. Answer the question posed in Part (d) if only 4200 of the students have both cards.

A single-elimination tournament with four players is to be held. In Game 1 , the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: \(P(\) seed 1 defeats seed 4\()=.8\) \(P(\) seed 1 defeats seed 2\()=.6\) \(P(\) seed 1 defeats seed 3\()=.7\) \(P(\) seed 2 defeats seed 3\()=.6\) \(P(\) seed 2 defeats seed 4\()=.7\) \(P(\) seed 3 defeats seed 4\()=.6\) a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate Game 3 in the tournament? (This will depend on the outcomes of Games 1 and 2.) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

Let \(F\) denote the event that a randomly selected registered voter in a certain city has signed a petition to recall the mayor. Also, let \(E\) denote the event that a randomly selected registered voter actually votes in the recall election. Describe the event \(E \cap F\) in words. If \(P(F)=.10\) and \(P(E \mid F)=.80\), determine \(P(E \cap F)\).

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