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✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()Unless otherwise noted, the coin flip has two disjoint outcomes—heads or tails—with probabilities
\[P(\text{heads}) = 0.5, \quad P(\text{tails}) = 0.5\] 
In the early formation of the field of statistics, there were considerations such as the following. How should the possibility space for a trial of flipping two coins be represented?
To sets are disjoint if \[P(A \cap B) = 0\]
A possibility space is a set of all of the possible outcomes for an event. It is usually denoted by the Greek letter capital omega.
For example, the set of all outcomes for two coin flips of a fair coin turns out to be
\[\Omega = \{HH, HT, TH, TT\}\]
If \(A\) is a set (and a subset of the possibility space), then the complement of A, denoted \(A^{c}\), is the set of outcomes that is in the universal set but not in the set A
\[A \subseteq \Omega \quad\Rightarrow\quad A^{c} = \Omega - A\]
For example, if we think of our roll of a six-sided die, the possibility space was \[\Omega = \{1, 2, 3, 4, 5, 6\}\] If we have a subset consisting of the even numbers \[E = \{2, 4, 6\}\] what do you think the complement \(E^{c}\) will be?
The empty set \(\{\}\) literally has zero elements in the set
Claim: Set \(A\) and its complement are disjoint.
There may be situations where we need to be careful about whether selections from a set were done with replacement or without replacement.
In the wardrobe, there are 8 blue socks and 6 red socks.
\[B, B, B, B, B, B, B, B\] \[R, R, R, R, R, R\]
Compute the following probabilities
Notice how when we sample with replacement, each iteration is independent of the previous iterations. When we sample without replacement, each iteration depends on the previous iterations.
One relationship between the notions of complements, intersections, and unions is as follows.
\[(A \cup B)^{c} = A^{c} \cap B^{c}\] The complement of the union is the intersection of the complements.
Exam 1 will be on Tues., Sept. 27
