Back to the on-screen lesson ·
Kolmogorov's three axioms, the rules derived from them, and counting when outcomes are equally likely.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
You will build a sample space for an experiment, say which subsets are the events you care about, state the three axioms, and derive from them the rules for complements, for subsets and for the union of two overlapping events. You will also say, of any rule about probabilities, whether the axioms alone give it or whether it needs an assumption they do not make.
You have met probability before as a fraction of outcomes, and you can add and subtract fractions. What is new is that the rules you already use are about to be derived rather than asserted, from three statements short enough to write on one line — so that the difference between a definition and a theorem stops being a matter of which one you happened to learn first.
The sample space $\Omega$ is the set of outcomes of one performance of an experiment. An event is a subset of it. Two events are disjoint when no outcome belongs to both. The complement $A^{c}$ is everything in $\Omega$ that is not in $A$. A probability measure is a function from events to numbers that obeys the three axioms below — and nothing else is assumed of it.
A sample space $\Omega$ lists the outcomes; an event is a subset. A probability $P$ assigns each event a number, subject to three axioms: $P(A) \ge 0$; $P(\Omega) = 1$; and $P(A_1 \cup A_2 \cup \cdots) = \sum_i P(A_i)$ whenever the $A_i$ are pairwise disjoint.
That is the whole list. Everything else this course states about probabilities in the abstract is derived from it: $P(\varnothing) = 0$, $P(A^{c}) = 1 - P(A)$, $P(A) \le P(B)$ when $A \subseteq B$, $P(A) \le 1$, and $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Knowing which statements are axioms and which are consequences is not pedantry: the two rules learners most often use as if they were axioms — that unions add and that intersections multiply — are true only under conditions, and the conditions are what the derivation makes visible.
When $\Omega$ is finite and every outcome is equally likely, $P(A) = |A| / |\Omega|$, and the subject becomes counting.
Another way: picture
A dartboard on which every point is equally likely: the probability of a region is its area divided by the board's. The axioms are exactly the properties area has — never negative, one in total, additive over pieces that do not overlap. Nothing about probability is stranger than area is.
Another way: steps
Each line uses only the axioms and the lines above it.
| Rule | Where it comes from |
|---|---|
| $P(\varnothing) = 0$ | $\varnothing$ is disjoint from itself, so $P(\varnothing) = P(\varnothing) + P(\varnothing)$ |
| $P(A^{c}) = 1 - P(A)$ | $A$ and $A^{c}$ are disjoint with union $\Omega$ |
| $A \subseteq B \Rightarrow P(A) \le P(B)$ | $B = A \cup (B \setminus A)$, disjointly, and the second piece is not negative |
| $P(A) \le 1$ | take $B = \Omega$ in the line above |
| $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | split $A \cup B$ into three disjoint pieces and add them two ways |
Read the table downwards and you can see the dependency: the last line cannot be proved before the first.
Treating additivity as unconditional. $P(A \cup B) = P(A) + P(B)$ is an axiom for disjoint events and false otherwise. Adding two overlapping events can produce a number above $1$, which is the clearest possible sign that a rule has been used outside its conditions.
Treating the product rule as an axiom. $P(A \cap B) = P(A)P(B)$ is the definition of independence. It is something you check, or assume out loud, never something the axioms hand you.
Assuming outcomes are equally likely because they can be listed. A sample space is a set; nothing about writing it down makes its members equally probable. Rain or no rain is a perfectly good two-element sample space and the two are not a half each.
Building a sample space whose outcomes overlap. The outcomes must be mutually exclusive and cover everything, or additivity is not available and none of the derived rules holds.
$A$ and $A^{c}$ are disjoint and their union is $\Omega$, so additivity gives $P(A) + P(A^{c}) = P(\Omega)$.
Axiom three, then axiom two.
$P(\Omega) = 1$, so $P(A^{c}) = 1 - P(A)$.
A theorem, not an axiom — and the proof is two lines.
It follows that $P(A) \le 1$, because $P(A^{c})$ is not negative.
Axiom one supplies the last step.
$\Omega$ is the $36$ ordered pairs, and the model says they are equally likely.
The model is an assumption; say it out loud.
Six pairs sum to $7$, so $P(\text{sum} = 7) = 6/36 = 1/6$.
Counting, because the outcomes are equally likely.
$P(\text{at least one six}) = 1 - P(\text{no six}) = 1 - 25/36 = 11/36$.
The complement is eleven cases shorter to count.
$P(A \cup B) = 0.6 + 0.5 - 0.3 = 0.8$.
Inclusion-exclusion, because the events overlap.
Neither is the complement of either: $P(A^{c} \cap B^{c}) = P((A \cup B)^{c})$.
De Morgan.
So the answer is $1 - 0.8 = 0.2$.
In a sample space of twenty equally likely outcomes, $A$ holds $4$ of them, $B$ holds $9$, and $2$ outcomes are in both. Fill in the three probabilities.
| Probability | |
|---|---|
| $P(A^{c})$ | |
| $P(A \cup B)$ | |
| $P(A^{c} \cap B^{c})$ |
Does $P(A \mid B) = P(B \mid A)$ for every pair follow from the three axioms alone?
One fair die, rolled once. Put these four events in order, least likely first.
Number the steps in order (write the number in the box):
$P(A) = 7/20$, $P(A \cup B) = 13/20$ and $P(A \cap B) = 1/20$. What is $P(B)$?
Answer:
$P(A) = 3/5$ and $P(B) = 3/5$, and nothing is said about how the two events overlap. Give the set of values $P(A \cap B)$ could take, as an interval.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An event has probability $3/10$. The marks are one twentieth apart. Put the marker at the probability of its complement.
0 |——————————| 1
Mark the position with a cross, then write the value:
You can write a sample space, state the three axioms and derive the complement, monotonicity and inclusion-exclusion rules from them. Say in your own words why adding the probabilities of two overlapping events gives the wrong answer, and what the correction is.
9. Your turn: $P(A) = 0.6$, $P(B) = 0.5$ and $P(A \cap B) = 0.3$; find the probability that neither happens, step 3