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Chapter 26: Problem 1

By shading Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove them using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\) (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \underline{(Y \cap Z)}=(X \cup Y) \cap Z\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\)

Short Answer

Expert verified
Relationships (a) and (e) are valid; (b) and (c) are invalid; (d) needs clarification or is invalid as written.

Step by step solution

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01

- Write Down Given Equations

Start by writing down all the given equations. We have five equations to check for validity: (a) \(\bar{(\bar{X} \cup Y)} = X \cap \bar{Y}\) (b) \(\bar{X} \cup \bar{Y} = \overline{(X \cup Y)}\) (c) \((X \cup Y) \cap Z = (X \cup Z) \cap Y\) (d) \(X \cup \underline{(Y \cap Z)} = (X \cup Y) \cap Z\) (e) \(X \cup \overline{(Y \cap Z)} = (X \cup \bar{Y}) \cup \bar{Z}\)
02

- Apply De Morgan's Laws to (a)

For (a), apply De Morgan's laws to the left side. According to De Morgan's laws, \(\overline{A \cup B} = \bar{A} \cap \bar{B}\). So, \(\bar{(\bar{X} \cup Y)}\) becomes \(\overline{\bar{X}} \cap \bar{Y}\). Since \(\overline{\bar{X}} = X\), we get \(X \cap \bar{Y}\). Therefore, relationship (a) is valid.
03

- Apply De Morgan's Laws to (b)

For (b), apply De Morgan's laws to the right side. According to De Morgan's laws, \(\overline{A \cup B} = \bar{A} \cap \bar{B}\). So, the right side \(\overline{(X \cup Y)}\) becomes \(\bar{X} \cap \bar{Y}\). Since the left side is \(\bar{X} \cup \bar{Y}\), which is not equal to \(\bar{X} \cap \bar{Y}\), relationship (b) is invalid.
04

- Simplify (c) Using Associative and Distributive Laws

For (c), simplify both sides using associative and distributive laws. The left side \((X \cup Y) \cap Z\) can be simplified as \((X \cap Z) \cup (Y \cap Z)\). The right side \((X \cup Z) \cap Y\) does not simplify to the same form as the left side because the union and intersection order is different. Therefore, relationship (c) is invalid.
05

- Simplify (d) Using Associative Laws

For (d), simplify both sides using associative laws. The left side \(X \cup (Y \cap Z)\) uses an incorrect notation with the underlining. Assuming we correct it to \((X \cup (Y \cap Z))\), it simplifies differently. Hence, as written, the relationship may not hold, and clarification is needed for the intent.
06

- Apply De Morgan's Laws to (e)

For (e), apply De Morgan's laws to the right side. According to De Morgan's laws, \(\overline{A \cap B} = \bar{A} \cup \bar{B}\). So, the left side \(X \cup \overline{(Y \cap Z)}\) becomes \(X \cup (\bar{Y} \cup \bar{Z})\) and using associative laws, we get \((X \cup \bar{Y}) \cup \bar{Z}\). Therefore, relationship (e) is valid.

Key Concepts

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

de Morgan's laws
De Morgan's laws are fundamental rules in set theory and logic, named after the mathematician Augustus De Morgan. These laws relate the complement of unions and intersections of sets. They are crucial for simplifying and proving set expressions. The two main laws are:
  • The complement of a union: \(\bar{(A \cup B)} = \bar{A} \cap \bar{B}\)
  • The complement of an intersection: \(\bar{(A \cap B)} = \bar{A} \cup \bar{B}\)
For example, in the problem provided, \(\bar{(\bar{X} \cup Y)} \) can be simplified using de Morgan's laws: First, de Morgan's law tells us to take the complement of each component: \(\bar{X} \) turns into \(\bar{X}\), and \( Y \) remains \(Y\). Then, replace the union with an intersection: \(\bar{(\bar{X} \cup Y)} \) becomes \(\bar{X} \cap \bar{Y}\). This helps in proving relationships effectively.
set operations
Set operations are the actions you can perform on sets, such as unions, intersections, and complements. These operations help us understand relationships between different sets. For clarity:
  • **Union (\(\cup\)**: Combines all elements from both sets. For example: \(A \cup B\) includes elements from either A or B, or both.
  • **Intersection (\(\cap\)**: Includes only elements common to both sets. For example: \(A \cap B\) includes only those elements found in both A and B.
  • **Complement (\(\bar{} \)**: Includes all elements not in the set. For example: \(\bar{A}\) includes all elements not in A.
For example: Consider the intersection \((X \cap Z)\). This includes only elements that are both in X and in Z. Understanding these operations is key to manipulating and simplifying complex set expressions.
logical equivalences
Logical equivalences are expressions that are true under the same conditions. These equivalences are used to simplify complex logical statements. Some important logical equivalences include:
  • **Identity Law**: \( A \cup \varnothing = A\) and \( A \cap U = A \), where \(U\) is the universal set and \(\backslash)varnothing\) is the empty set.
  • **Domination Law**: \(A \cup U = U\) and \(A \cap \varnothing = \varnothing\)
  • **Idempotent Law**: \(A \cup A = A\) and \(A \cap A = A\)
  • **Double Negation Law**: \(\bar{\bar{A}} = A\)
For instance, in the exercise given, de Morgan's laws help verify relationships by transforming the complement and intersection operations into simpler equivalent forms.
associative laws
The associative laws allow us to regroup set operations without changing the result. These laws are especially useful for simplifying and comparing set expressions. The associative laws are:
  • **Associative Law for Union**: \( (A \cup B) \cup C = A \cup (B \cup C)\)
  • **Associative Law for Intersection**: \( (A \cap B) \cap C = A \cap (B \cap C)\)
For example, \((X \cup Y) \cap Z\) in one of the relationships can be simplified using associative laws. Rewriting the expression can show if two sides of an equation are equivalent. This helps in understanding complex relationships among sets by breaking them down into simpler parts.

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Most popular questions from this chapter

Villages \(A, B, C\) and \(D\) are connected by overhead telephone lines joining \(A B\), \(A C, B C, B D\) and \(C D .\) As a result of severe gales, there is a probability \(p\) (the same for each link) that any particular link is broken. (a) Show that the probability that a call can be made from \(A\) to \(B\) is $$ 1-2 p^{2}+p^{3} $$ (b) Show that the probability that a call can be made from \(D\) to \(A\) is $$ 1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} $$

\(26.3 A\) and \(B\) each have two unbiased four-faced dice, the four faces being numbered \(1,2,3,4\). Without looking, \(B\) tries to guess the sum \(x\) of the numbers on the bottom faces of \(A\) 's two dice after they have been thrown onto a table. If the guess is correct \(B\) receives \(x^{2}\) euros, but if not he loses \(x\) euros. Determine \(B\) 's expected gain per throw of \(A\) 's dice when he adopts each of the following strategies: (a) he selects \(x\) at random in the range \(2 \leq x \leq 8\); (b) he throws his own two dice and guesses \(x\) to be whatever they indicate; (c) he takes your advice and always chooses the same value for \(x\). Which number would you advise? 26.4 Use the method of induction to prove equation (26.16), the probability addition law for the union of \(n\) general events.

A continuous random variable \(X\) is uniformly distributed over the interval \([-c, c]\). A sample of \(2 n+1\) values of \(X\) is selected at random and the random variable \(Z\) is defined as the median of that sample. Show that \(Z\) is distributed over \([-c, c]\) with probability density function, $$ f_{n}(z)=\frac{(2 n+1) !}{(n !)^{2}(2 c)^{2 n+1}}\left(c^{2}-z^{2}\right)^{n} $$ Find the variance of \(Z\).

Two continuous random variables \(X\) and \(Y\) have a joint probability distribution $$ f(x, y)=A\left(x^{2}+y^{2}\right) $$ where \(A\) is a constant and \(0 \leq x \leq a, 0 \leq y \leq a\). Show that \(X\) and \(Y\) are negatively correlated with correlation coefficient \(-15 / 73 .\) By sketching a rough contour map of \(f(x, y)\) and marking off the regions of positive and negative correlation, convince yourself that this (perhaps counter-intuitive) result is plausible.

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.
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