Probability — chapter notes

A random experiment has several possible results and the result cannot be predicted in advance. Each result is an outcome, and the set of all outcomes is the sample space $S$.

Write $S$ as a set and state $n(S)$. For a two-stage experiment follow each branch. Decide whether order matters and whether items are replaced: choosing $2$ from $4$ gives $16$ outcomes with replacement and order, $12$ without replacement but in order, and $6$ when the two are picked together.

Outcomes need not be equally likely; the sample space still lists all of them.

An event is a subset of the sample space. An event occurs when the outcome is one of its members. A simple (elementary) event has one outcome and a compound event has more. The whole space $S$ is the sure event; the empty set $\varnothing$ is the impossible event.

If $S$ has $n$ outcomes there are $2^n$ events, counting $S$ and $\varnothing$. For a spinner numbered $1$ to $10$, "even" is $\{2, 4, 6, 8, 10\}$ and "greater than $10$" is $\varnothing$.

With events $A$ and $B$ of the same sample space: $A$ or $B$ is $A \cup B$ (at least one occurs); $A$ and $B$ is $A \cap B$; not $A$ is $A'$; $A$ but not $B$ is $A - B = A \cap B'$; neither is $A' \cap B' = (A \cup B)'$.

A Venn diagram with the sample space as a rectangle shows every region. For the even numbers and the multiples of $3$ on a spinner numbered $1$ to $10$: $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$, $A \cap B = \{6\}$ and the neither-event is $\{1, 5, 7\}$.

$A$ and $B$ are mutually exclusive if $A \cap B = \varnothing$: they cannot occur together. They are exhaustive if $A \cup B = S$: at least one must occur. Both together means each is the complement of the other.

Events $E_1, E_2, \ldots, E_n$ that are pairwise mutually exclusive and together exhaustive form a partition of $S$. Dividing the numbers $1$ to $12$ by $3$ gives such a partition by remainders $0, 1, 2$.

To classify a pair, find the intersection (empty?) and the union (equal to $S$?).

A probability $P$ assigns a number to each event so that: (1) $P(E) \ge 0$; (2) $P(S) = 1$; (3) $P(A \cup B) = P(A) + P(B)$ whenever $A \cap B = \varnothing$.

To define a model, give each elementary event a probability: all non-negative, total $1$. The probability of an event is the sum of the probabilities of its outcomes. If an unknown such as $x$ appears, make the total equal to $1$ first.

A proposed list is valid exactly when no entry is negative and the total is $1$.

If all $n(S)$ outcomes are equally likely, each has probability $\dfrac{1}{n(S)}$ and $P(E) = \dfrac{n(E)}{n(S)}$. This is a special case of the axioms.

From the axioms: $P(\varnothing) = 0$; $P(A') = 1 - P(A)$ (since $A$, $A'$ are exclusive with union $S$); and $0 \le P(E) \le 1$. Use the complement when "at least one" is hard to count: for four coin tosses $P(\text{at least one head}) = 1 - \dfrac{1}{16} = \dfrac{15}{16}$.

For any two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. The outcomes of $A \cap B$ are in both $P(A)$ and $P(B)$, so the overlap is subtracted once.

Related results: $P(\text{neither}) = 1 - P(A \cup B)$; $P(A \cap B') = P(A) - P(A \cap B)$; $P(\text{exactly one}) = P(A \cup B) - P(A \cap B)$. For mutually exclusive events the overlap is $0$, so $P(A \cup B) = P(A) + P(B)$; if they are also exhaustive the two probabilities add to $1$.

For a set of equally likely outcomes, verify by counting: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

First decide whether the outcomes are equally likely: if yes, count $n(E)$ and $n(S)$; if not, use the given or weighted probabilities and add. Then pick the rule for the event wanted: not $A$ uses the complement; $A$ or $B$ uses the addition theorem; neither, $A$ but not $B$ and exactly one all come from $P(A \cup B)$ and $P(A \cap B)$.

Always check that every probability is between $0$ and $1$, and that exclusive, exhaustive events have probabilities adding to $1$.