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Let \(A\) and \(B\) be two events such that \(P\left((A \cup B)^{C}\right)=\) \(0.6\) and \(P(A \cap B)=0.1\). Let \(E\) be the event that either \(A\) or \(B\) but not both will occur. Find \(P(E \mid A \cup B)\).

Short Answer

Expert verified
The probability \(P(E \mid A \cup B)\) is more than 0.25

Step by step solution

01

Understand Event E

The event \(E\) is that either \(A\) or \(B\) but not both will occur. This means \(E = (A \cap B^{C}) \cup (A^{C} \cap B)\). This event accounts for the occurrence of either event \(A\) or \(B\) but not both.
02

Apply the formula of conditional probability

The formula for conditional probability is given as \(P(A∣B) = P(A \cap B) / P(B)\). Combining it with the definition of \(E\) we got in step 1, we now have \(P(E \mid A \cup B) = P(E \cap (A \cup B)) / P(A \cup B)\).
03

Calculate P(A∪B) and P(E∩(A∪B))

We already know that the probability of the union of two events is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). We need to find \(P(A)\) and \(P(B)\). From given data, we have \(P((A \cup B)^{C}) = 0.6\), therefore \(P(A \cup B) = 1 - 0.6 = 0.4\). Calculating \(P(A)\) and \(P(B)\) using these values might be complex, but we can find \(P(E \cap (A \cup B))\) without actually finding individual values for \(P(A)\) or \(P(B)\). From the definition of \(E\) in step 1, we see that \(E \cap (A \cup B)\) equals \(E\). Hence, \(P(E \cap (A \cup B)) = P(E)\).
04

Calculate P(E)

We have \(P(E) = P((A \cap B^{C}) \cup (A^{C} \cap B)) = P(A \cap B^{C}) + P(A^{C} \cap B) - P((A \cap B^{C}) \cap (A^{C} \cap B))\). The third term on the right becomes 0 as \(A\) cannot simultaneously be true and false, same case for \(B\). Therefore, \(P(E) = P(A \cap B^{C}) + P(A^{C} \cap B)\). With given data, we see that \(P(A \cap B^{C})\) and \(P(A^{C} \cap B)\) will both be greater than \(P(A \cap B) = 0.1\). Therefore \(P(E) > 0.1\).
05

Calculate the final result

From step 2, we know that \(P(E \mid A \cup B) = P(E \cap (A \cup B)) / P(A \cup B) = P(E) / 0.4\). From step 4, we know that \(P(E) > 0.1\). Hence, \(P(E \mid A \cup B) > 0.1 / 0.4 = 0.25\)

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Key Concepts

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

Probability Theory
Probability theory is a fundamental part of mathematics that deals with the likelihood of events occurring. It provides a framework for reasoning about uncertainty and making predictions based on incomplete information. The key principle here is that the probability of any event ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty. Probability theory is used in various fields, from statistics and sciences to social sciences and finance.
To understand probability, we often deal with random experiments, which are processes that lead to uncertain outcomes. The set of all possible outcomes is called the sample space, and each possible outcome is referred to as an event. We assign probabilities to these events, determining how likely each one is to occur. These probabilities must sum to 1, meaning one of the outcomes must happen.
Conditional probability, as shown in the exercise solution, is a concept where we determine the probability of an event occurring given that another event has already occurred. It is represented as \(P(A \mid B)\), read as 'the probability of event A given event B', and calculated with the formula \(P(A \mid B) = P(A \cap B) / P(B)\). This concept helps us refine probabilities based on new information.
Union and Intersection of Events
In probability theory, understanding the union and intersection of events is vital. These operations help describe complex event combinations using simpler constituent events.
The union of events, denoted as \(A \cup B\), refers to the occurrence of at least one of the events A or B. The probability of the union of two events is given by the formula
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula ensures we don't double-count the probability of both events occurring together, represented by the intersection.
"Intersection," denoted as \(A \cap B\), is when both events A and B occur simultaneously. In terms of probability, it quantifies the likelihood of two events happening at the same time. For instance, if you want to find the probability that a card drawn is both an ace and red from a deck, you'd calculate the intersection of the events "being an ace" and "being red."
Using these foundational concepts, the exercise task can decompose a complex event (E) into simpler intersections and unions, making it easier to calculate probabilities.
Complementary Events
Complementary events are another key idea in probability theory. They help us understand the 'opposite' or 'rest' of an event. For any event A, the complementary event, denoted by \(A^C\), includes everything in the sample space that does not belong to A. A practical example is flipping a coin, where if event A is 'landing on heads', then \(A^C\) is 'landing on tails'.
The relationship between an event and its complement is crucial because their probabilities add up to 1:
  • \(P(A) + P(A^C) = 1\)
This tells us that, in any situation, either the event happens, or it doesn't. Knowing either probability allows us to find the other. This can simplify calculations, such as finding \(P(A^C)\) by using \(1 - P(A)\).
In solving the exercise, the complement of the union of events \((A \cup B)^C\) is used to find the probability of \(A \cup B\). By establishing the probability of the complement, the probability of the primary event is easily calculated, showcasing how complementary events can make complex probability problems more digestible. This approach highlights the power of complement in breaking down probability dilemmas.

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Most popular questions from this chapter

A fair die is rolled and then \(n\) fair coins are tossed, where \(n\) is the number showing on the die. What is the probability that no heads appear?

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