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Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F)\) ? b. Suppose that for events \(A\) and \(B, P(A)=0.24, P(B)=0.24\), and \(P(A \cup B)=0.48 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Short Answer

Expert verified
i. The value of \(P(E \cap F)\) is 0. ii. The value of \(P(E \cup F)\) is 0.9. b. Yes, events A and B are mutually exclusive.

Step by step solution

01

Compute the Intersection of Mutually Exclusive Events

For mutually exclusive events, the intersection, represented by \(P(E \cap F)\), is always zero because mutually exclusive events cannot occur at the same time. Since E and F are mutually exclusive: \(P(E \cap F)=0\). This is always true and no computation is needed.
02

Compute the Union of Mutually Exclusive Events

For mutually exclusive events, the probability of their union is the sum of their individual probabilities. So, \(P(E \cup F)\) = \(P(E)\) + \(P(F) = 0.14 + 0.76 = 0.9\).
03

Check if Events A and B are Mutually Exclusive

Two events A and B are mutually exclusive if the probability of their union is equal to the sum of their individual probabilities. Therefore, if \(P(A \cup B) = P(A) + P(B)\), then A and B are mutually exclusive. Let's see if that is true. Given values are: \(P(A) = 0.24\), \(P(B) = 0.24\), and \(P(A \cup B) = 0.48\). From calculation, \(P(A) + P(B) = 0.24 + 0.24 = 0.48\). As this equals to \(P(A \cup B)\), we conclude A and B are mutually exclusive.

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events are a key concept in probability theory. They refer to events that cannot happen at the same time. If one event occurs, the other cannot. An easy way to remember this is by thinking of events that completely cancel each other out.
For example, if you have events E and F that are mutually exclusive, it means if E happens, F cannot, and vice versa. Mathematically, this is represented as their probability of intersection being zero:
  • \(P(E \cap F) = 0\)
To identify mutually exclusive events, think about scenarios where the occurrence of one event simultaneously makes the other's occurrence impossible. This concept is crucial, especially when solving problems that involve evaluating the probability of either event happening.
Probability of Union
The probability of the union of two events is about finding the likelihood that at least one of the events occurs. In mathematics, this is expressed as \(P(E \cup F)\), indicating either event E or event F happens, or both.
For mutually exclusive events, the probability of their union simplifies to the sum of their individual probabilities:
  • \(P(E \cup F) = P(E) + P(F)\)
This formula works simply because, for mutually exclusive events, both events cannot occur together. So, when we add their probabilities, we do not double-count any outcome. As an example, if \(P(E) = 0.14\) and \(P(F) = 0.76\), then \(P(E \cup F) = 0.14 + 0.76 = 0.9\).
However, when dealing with non-mutually exclusive events, this equation needs adjustment as it would overestimate the probability by including the intersection twice. Thus, a correction term is used, adjusting for the overlapping probability.
Probability of Intersection
The concept of the probability of intersection focuses on finding the likelihood that two events happen simultaneously. This is represented by \(P(E \cap F)\) in probability theory. For mutually exclusive events, this probability is zero because they cannot occur together. Hence, \(P(E \cap F) = 0\).
However, when we deal with non-mutually exclusive events, the probability of intersection becomes a critical part of calculating the union of the two events. It's used to adjust the sum of their individual probabilities by subtracting this joint probability, ensuring no overlapping probability is counted twice. This is seen in the relationship:
  • \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)
Understanding the probability of intersection is vital, especially in problems where events may overlap or influence each other. It helps in accurately assessing the overall probability of combined events.

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Most popular questions from this chapter

An article in the New York Times 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 \()=0.01\) for people who suffer cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and difficulty finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The basis for the New York Times article was a research study of 2,329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2,329 cardiac arrest sufferers do you think survived? Explain.

A small college has 2,700 students enrolled. Consider the chance experiment of selecting a student at random. For each of the following pairs of events, indicate whether or not you think they are mutually exclusive and explain your reasoning. a. the event that the selected student is a senior and the event that the selected student is majoring in computer science. b. the event that the selected student is female and the event that the selected student is majoring in computer science. c. the event that the selected student's college residence is more than 10 miles from campus and the event that the selected student lives in a college dormitory. d. the event that the selected student is female and the event that the selected student is on the college football team.

The authors of the paper "Do Physicians Know when Their Diagnoses Are Correct?" (Journal of General Internal Medicine [2005]: \(334-339\) ) presented detailed case studies to students and faculty at medical schools. Each participant was asked to provide a diagnosis in the case and also to indicate whether his or her confidence in the correctness of the diagnosis was high or low. Define the events \(C, I,\) and \(H\) as follows: \(C=\) event that diagnosis is correct \(I=\) event that diagnosis is incorrect \(H=\) event that confidence in the correctness of the diagnosis is high a. Data appearing in the paper were used to estimate the following probabilities for medical students: $$\begin{aligned} P(C) &=0.261 & & P(I)=0.739 \\ P(H \mid C) &=0.375 & & P(H \mid I)=0.073 \end{aligned} $$Use the given probabilities to construct a "hypothetical 1000 " table with rows corresponding to whether the diagnosis was correct or incorrect and columns corresponding to whether confidence was high or low. b. Use the table to calculate the probability of a correct diagnosis, given that the student's confidence level in the correctness of the diagnosis is high. c. Data from the paper were also used to estimate the following probabilities for medical school faculty: $$\begin{array}{cl} P(C)=0.495 & P(I)=0.505 \\ P(H \mid C)=0.537 & P(H \mid I)=0.252 \end{array}$$ Construct a "hypothetical \(1000 "\) ' table for medical school faculty and use it to calculate the probability of a correct diagnosis given that the faculty member's confidence level in the correctness of the diagnosis is high. How does the value of this probability compare to the value for students calculated in Part (b)?

An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is \(0.86 .\) Give a relative frequency interpretation of this probability.

What does it mean to say that the probability that a coin toss will land head side up is \(0.5 ?\)

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