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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

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.

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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

\(26.3 A\) 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 shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the final digit (the number of pence) in the price is uniformly and randomly distributed from 0 to \(9 .\) Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probability if the price ends in a ' 5 '. Should she suspect a mistake if the cashier asks her for 23 pence more than she estimated?

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 $$ 1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} $$

(a) In two sets of binomial trials \(T\) and \(t\) the probabilities that a trial has a successful outcome are \(P\) and \(p\) respectively, with corresponding probabilites of failure of \(Q=1-P\) and \(q=1-p .\) One 'game' consists of a trial \(T\) followed, if \(T\) is successful, by a trial \(t\) and then a further trial \(T .\) The two trials continue to alternate until one of the \(T\) trials fails, at which point the game ends. The score \(S\) for the game is the total number of successes in the t-trials. Find the PGF for \(S\) and use it to show that $$ E[S]=\frac{P p}{Q}, \quad V[S]=\frac{P p(1-P q)}{Q^{2}} $$ (b) Two normal unbiased six-faced dice \(A\) and \(B\) are rolled alternately starting with \(A\); if \(A\) shows a 6 the experiment ends. If \(B\) shows an odd number no points are scored, if it shows a 2 or a 4 then one point is scored, whilst if it records a 6 then two points are awarded. Find the average and standard deviation of the score for the experiment and show that the latter is the greater.

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n} .\) A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all its possible values with equal probability and hence demonstrate that two random variables can be uncorrelated even though one is defined in terms of the other.
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