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Chapter 7: Problem 45

Let \(S\) be a sample space for an experiment, and let \(E\) and \(F\) be events of this experiment. Show that the events \(E \cup F\) and \(E^{c} \cap F^{c}\) are mutually exclusive. Hint: Use De Morgan's law.

Short Answer

Expert verified
Using De Morgan's law \((A \cup B)^c = A^c \cap B^c\), we rewrite the given events and apply the distributive law to find their intersection: \((E \cup F) \cap (E^c \cap F^c) = (E \cap E^c) \cup (F \cap F^c)\). Since \(E \cap E^c = \emptyset\) and \(F \cap F^c = \emptyset\), the intersection is \(\emptyset\), proving that the events \(E \cup F\) and \(E^c \cap F^c\) are mutually exclusive.

Step by step solution

01

Define mutually exclusive events

Mutually exclusive events are events that cannot both occur at the same time. Mathematically, if E and F are mutually exclusive events, then their intersection is an empty set, i.e., \(E \cap F = \emptyset\).
02

Write the given events using set notation

We're given the events \(E \cup F\) and \(E^c \cap F^c\). Using set notation, we can write these events as follows: - Event \(E \cup F\) is the union of sets E and F. - Event \(E^c \cap F^c\) is the intersection of the complements of sets E and F.
03

Apply De Morgan's law

We will use De Morgan's law to rewrite the event \(E^c \cap F^c\). De Morgan's law states that: \((A \cup B)^c = A^c \cap B^c\). In our case, we will set A = \(E^c\) and B = \(F^c\), and then apply the law to obtain: \((E^c \cap F^c)^c = (E^c)^c \cup (F^c)^c\), which simplifies to \(E \cup F = E^c \cap F^c\).
04

Determine the intersection of the events

To show that the events \(E \cup F\) and \(E^c \cap F^c\) are mutually exclusive, we need to determine their intersection: \((E \cup F) \cap (E^c \cap F^c)\).
05

Apply the distributive law

We will use the distributive law to find the intersection: \((E \cup F) \cap (E^c \cap F^c)\). The distributive law states that \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\). By setting A = \(E \cup F\), B = \(E^c\), and C = \(F^c\), we obtain the following: \((E \cup F) \cap (E^c \cap F^c) = (E \cap E^c) \cup (F \cap F^c)\).
06

Evaluate the intersection terms

Now, we need to evaluate the intersection terms \((E \cap E^c)\) and \((F \cap F^c)\). Given that \(E^c\) and \(F^c\) are the complements of sets E and F, their intersections with their corresponding sets will always be empty sets or \(E \cap E^c = \emptyset\) and \(F \cap F^c = \emptyset\).
07

Prove the events are mutually exclusive

Due to the intersection terms being empty sets, we can rewrite step 5 as follows: \((E \cup F) \cap (E^c \cap F^c) = \emptyset \cup \emptyset = \emptyset\). Since the intersection of events \(E \cup F\) and \(E^c \cap F^c\) is an empty set, we can now conclude that these events are mutually exclusive.

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