Chapter 18: Problem 792
\(P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8\) \(P(A \cap C)=0.3, P(A \cap B \cap C)=0.2\) and \(P(A \cup B \cup C) \geq 0.85\) Then range of \(P(B \cap C)\) is (a) \([0.3,0.4]\) (b) \([0.1,0.25]\) (c) \([0.2,0.35]\) (d) None
Short Answer
Expert verified
The range for \(P(B \cap C)\) is \([0, 0.35]\), which doesn't match with any of the given options, hence the answer is (d) None.
Step by step solution
01
The principle of inclusion and exclusion refers to a method for counting the size of a union of several sets. For three events A, B, and C it can be written as: \[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\] #Step 2: Substitute given probabilities in the equation#
We have the following information: \(P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=0.8, P(A \cap C)=0.3, P(A \cap B \cap C)=0.2\) and \(P(A \cup B \cup C) \geq 0.85\). We can substitute these probabilities into the principle of inclusion and exclusion equation:
\[P(A \cup B \cup C) = 0.6 + 0.4 + 0.5 - P(A \cap B) - 0.3 - P(B \cap C) + 0.2\]
We also know that \(P(A \cup B) = 0.8\), so we can use this information to find \(P(A \cap B)\).
#Step 3: Find \(P(A \cap B)\)#
02
Since \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), we can solve for \(P(A \cap B)\) as follows: \[0.8 = 0.6 + 0.4 - P(A \cap B)\] \[P(A \cap B) = 0.6 + 0.4 - 0.8 = 0.2\] #Step 4: Determine the range of \(P(B \cap C)\)#
Now substitute the value of \(P(A \cap B)\) back in the equation of step 2, and we have:
\[P(A \cup B \cup C) \geq 0.85 = 0.6 + 0.4 + 0.5 - 0.2 - 0.3 - P(B \cap C) + 0.2\]
Simplifying, we get:
\[0.85 - 1.2 \leq - P(B \cap C)\]
\[-0.35 \leq - P(B \cap C)\]
\[0.35 \geq P(B \cap C)\]
From this inequality we can tell that the upper bound for \(P(B \cap C)\) is 0.35. Now we will find its lower bound. Since the intersection of any two events cannot be negative, the smallest value for \(P(B \cap C)\) is when it equals zero, in which case the events B and C would be mutually exclusive.
Since \(P(B \cap C) \geq 0\), we have found our lower bound.
The range for \(P(B \cap C)\) is thus, \(\boxed{[0, 0.35]}\), which doesn't match with any of the given options, hence the answer is (d) None.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Principle of Inclusion and Exclusion
Understanding the principle of inclusion and exclusion is essential for solving complex probability problems involving multiple events. This principle is based on the idea of compensating for over-counting when calculating the probability of the union of multiple sets or events. When we simply add the probabilities of each event, we count the probability of their intersections more than once.
In terms of set theory, when we want to find the probability of the event that at least one of several events occurs, we start by adding the probabilities of each individual event. Then, we subtract the probabilities of the pairwise intersections between events, as these were counted multiple times. However, by subtracting these pairwise intersections, we remove the probability of the intersection of all three events entirely, so we must add this back.
In terms of set theory, when we want to find the probability of the event that at least one of several events occurs, we start by adding the probabilities of each individual event. Then, we subtract the probabilities of the pairwise intersections between events, as these were counted multiple times. However, by subtracting these pairwise intersections, we remove the probability of the intersection of all three events entirely, so we must add this back.
Application in Exercises
In the exercise given, the principle of inclusion and exclusion is used to find the probability of the union of three sets A, B, and C. It is crucial to apply this principle carefully to avoid mistakes in the calculation. When substituting the values into the formula, double-checking the work is advisable to ensure each term is accurately accounted for. The step-by-step solution clearly shows how the principle is applied to determine the probability of the union of the three given events.Set Theory in Probability
Set theory is the mathematical language of probability. It provides a framework for discussing collections of objects, known as sets, and the relationships between them. In probability, each event can be represented as a set containing all outcomes that result in that event. The fundamental operations in set theory—such as union, intersection, and complement—are directly relevant to calculating probabilities of events.
In the context of the problem at hand, events A, B, and C can be seen as sets with certain probabilities. The probabilities of these sets reflect the likelihood of each event occurring. Set theory in probability allows us to make meaningful statements about these probabilities and to use operations on sets to find probabilities of combined events, such as 'A or B' or 'A and B'.
In the context of the problem at hand, events A, B, and C can be seen as sets with certain probabilities. The probabilities of these sets reflect the likelihood of each event occurring. Set theory in probability allows us to make meaningful statements about these probabilities and to use operations on sets to find probabilities of combined events, such as 'A or B' or 'A and B'.
Visualizing with Venn Diagrams
Venn diagrams are a helpful visual tool to represent set operations in probability. These diagrams can simplify the understanding of complex relationships between events, such as overlapping areas indicating the intersection of sets. It can be a useful exercise enhancement to draw these diagrams when solving probability problems, as they can make abstract concepts much more tangible.Union and Intersection of Sets
The union and intersection of sets are fundamental operations that help solve a wide range of probability problems. The union of two or more sets is the set of elements that belong to at least one of the sets. In probability terms, it corresponds to the occurrence of at least one of the events represented by those sets. For the union, we use the symbol \(\cup\).
The intersection of sets, denoted by \(\cap\), represents a set containing elements that are common to all the involved sets. In probability, this means the intersection represents the event where all of the events occur simultaneously. It is crucial to understand the distinction between these two operations when solving problems that involve multiple events.
The intersection of sets, denoted by \(\cap\), represents a set containing elements that are common to all the involved sets. In probability, this means the intersection represents the event where all of the events occur simultaneously. It is crucial to understand the distinction between these two operations when solving problems that involve multiple events.