Probability · Lesson 3 of 8
Probability Axioms and Outcome Weights
“Build a valid probability model and find an event’s chance by adding the weights of its outcomes.”
• State and interpret the three probability axioms for finite sample spaces. • Explain why the empty event has probability zero. • Check that outcome weights are nonnegative and total one. • Calculate probabilities by summing the appropriate outcome weights. • Distinguish equally weighted outcomes from unequal probability assignments.
Sets tell us what; probabilities tell us how likely
An event tells us which results satisfy a condition, but the set alone does not always tell us their chances. A coin with a tendency towards heads still has the two outcomes H and T. To describe that tendency, we attach different numerical weights to the two outcomes. Probability gives a consistent way to attach and combine such weights.
A probability is a number between 0 and 1. Zero represents no probability weight and one represents the full weight of the experiment. Values such as 1/4 or 0.25 represent one quarter of that total weight. Writing 25% is another expression of the same value; it is not a different model.
A rule P that assigns a number P(E) to every event E in the sample space, obeying the probability axioms.
The domain of P is the collection of all subsets of S, called the power set of S. For S={H,T}, that collection is {∅,{H},{T},S}. Thus P acts on events, not just on the symbols used for individual results. Once we have consistent probabilities for the simple events of a finite S, we can obtain the probability of any event by adding its member weights.
Three rules that keep the model consistent
The axiomatic approach begins with rules that a valid probability assignment must obey. They express that chance cannot be negative, that some outcome of the stated sample space must occur, and that separate alternatives can have their weights added without double counting.
The third axiom does not say that all unions have probabilities given by a simple sum. If E and F share outcomes, those outcomes would be included in both P(E) and P(F). We will handle that overlap later. Repeated use of the axiom does allow the sum of any finite collection of pairwise disjoint events.
Problem
Derive P(∅)=0 using the axioms.
- 1.Take any event E. The sets E and ∅ are disjoint, and E∪∅=E.
- 2.Disjoint additivity gives P(E∪∅)=P(E)+P(∅).
- 3.Replace E∪∅ by E to obtain P(E)=P(E)+P(∅).
- 4.Subtract P(E) from both sides. The only possible value is P(∅)=0.
No event can have weight above 1. To see this, split S into E and the outcomes outside E. These sets are disjoint, and the total weight 1 is the sum of their nonnegative weights. Neither part can exceed that total. Together with nonnegativity, this gives 0≤P(E)≤1.
Assigning weights to individual outcomes
Suppose S={ω₁,…,ωₙ} is finite. Each singleton {ωᵢ} is a simple event, and different singletons are disjoint. The whole sample space is their union, so its weight must be their sum. This gives two practical tests for an assignment: no negative weight, and a total of exactly one.
For convenience, P(ωᵢ) often means P({ωᵢ}). The input is still the singleton event; the braces have merely been suppressed.
For one coin toss, P(H)=1/2 and P(T)=1/2 is a valid assignment. P(H)=1/4 and P(T)=3/4 is also valid: both weights lie in the allowed interval and sum to one. More generally, any p with 0≤p≤1 gives the valid pair p,1−p. The axioms specify consistency, while information about the coin is needed to decide which model describes it.
The model alone does not promise that a short run of tosses will reproduce its probabilities as exact proportions. An uncertain experiment can give an uneven short sequence even when the model uses equal weights. Conversely, counting two possible labels is not evidence that the coin is fair.
| Assignment for six outcomes | ω₁ | ω₂ | ω₃ | ω₄ | ω₅ | ω₆ |
|---|---|---|---|---|---|---|
| a | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
| b | 1 | 0 | 0 | 0 | 0 | 0 |
| c | 1/8 | 2/3 | 1/3 | 1/3 | −1/4 | −1/3 |
| d | 1/12 | 1/12 | 1/6 | 1/6 | 1/6 | 3/2 |
| e | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
Problem
Which rows of the six-outcome table are valid probability assignments?
- 1.Row a is valid: every entry is 1/6 and six such entries total 1.
- 2.Row b is also valid: 1+0+0+0+0+0=1, and all entries are in [0,1]. The axioms allow endpoint values.
- 3.Row c is invalid because two entries are negative. A favourable total would not repair negative probabilities.
- 4.Row d is invalid because 3/2 is greater than 1.
- 5.Row e is invalid because its total is 2.1 rather than 1, even though each individual entry is between 0 and 1.
Checking individual bounds is not enough. Six numbers can each lie between 0 and 1 while their total exceeds 1. Checking only the total is also insufficient if negative entries are present.
Adding the weights inside an event
An event is the union of the simple events corresponding to its members. Since those simple events are disjoint, add their weights. This applies whether the weights are equal or unequal, and it prevents us from using a counting fraction when equality has not been established.
| Pen sequence | BBB | BBG | BGB | GBB | BGG | GBG | GGB | GGG |
|---|---|---|---|---|---|---|---|---|
| Assigned weight | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 | 1/8 |
Problem
Three consecutive pens are classified as bad B or good G. Using the eight equal weights in the table, find the chances of exactly one bad pen and at least two bad pens.
- 1.Exactly one bad pen corresponds to A={BGG,GBG,GGB}.
- 2.P(A)=P(BGG)+P(GBG)+P(GGB)=1/8+1/8+1/8=3/8.
- 3.At least two bad pens corresponds to B={BBG,BGB,GBB,BBB}.
- 4.P(B)=1/8+1/8+1/8+1/8=4/8=1/2. The probabilities here come from the supplied model, not merely from the letters good and bad.
| Two-toss outcome | HH | HT | TH | TT |
|---|---|---|---|---|
| Assigned probability | 1/4 | 1/7 | 2/7 | 9/28 |
Problem
Using the two-toss table, find the probability of equal results and of exactly two heads.
- 1.Check the model: 1/4+1/7+2/7+9/28=(7+4+8+9)/28=1.
- 2.Equal results form E={HH,TT}, so P(E)=1/4+9/28=7/28+9/28=16/28=4/7.
- 3.Exactly two heads forms F={HH}, so P(F)=1/4.
- 4.Counting two outcomes out of four would give 1/2 for E, which is wrong here because the four outcomes do not have equal weights.
Problem
A die has two faces labelled 1, three labelled 2, and one labelled 3. Assuming its six physical faces are equally likely, find P(2), P(1 or 3), and P(not 3).
- 1.There are six equally weighted physical faces. Combining faces with the same number gives P(1)=2/6, P(2)=3/6, P(3)=1/6.
- 2.Thus P(2)=1/2. The recorded numbers 1,2,3 are not equally likely.
- 3.P(1 or 3)=2/6+1/6=1/2 because the two recorded-number events are disjoint.
- 4.Not 3 consists of the faces labelled 1 or 2, so its weight is 2/6+3/6=5/6.
A nonempty event can have probability zero in an assigned finite model; row b provides such examples. The impossible event is specifically the empty set. Similarly, a model may attach total weight one to an event smaller than S. Keep the set definitions separate from the numerical weights.
The next table provides seven-outcome assignments for practice. Apply the two model tests separately: inspect the entries, then calculate their total. Each row is a separate proposed model for the same seven sample points.
| Assignment | ω₁ | ω₂ | ω₃ | ω₄ | ω₅ | ω₆ | ω₇ |
|---|---|---|---|---|---|---|---|
| a | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| b | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 |
| c | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| d | −0.1 | 0.2 | 0.3 | 0.4 | −0.2 | 0.1 | 0.3 |
| e | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |
Quiz
Which assignment to three outcomes is valid?
Which condition permits P(E∪F)=P(E)+P(F) directly?
Why is P(∅)=0?
In the unequal two-toss model, what is P({HT,TH})?
Which is allowed by the axioms for two outcome weights?
A model has weights 0.1,0.2,0.3,0.4. An event contains the second and fourth outcomes. What is its probability?
Practice Problems
- For the seven-outcome assignment table, decide which rows a–e are valid and give a specific reason for rejecting each invalid row. Check the individual weights and their sum.
- A die has two faces labelled 1, three labelled 2, and one labelled 3. For equally likely physical faces, find P(2), P(1 or 3), and P(not 3). Explain why assigning 1/3 to each recorded number is incorrect.
- For S={a,b,c,d}, weights are 0.15,0.25,0.10,x. Find x, then find the probabilities of {a,c} and {b,d}.
- Give two different valid assignments to S={H,T}. Explain what the axioms establish and what information about the coin is still needed.
- Using the supplied three-pen model, find the probability of no bad pen and of exactly two bad pens. Explain why three consecutive pens need not have these probabilities in every real machine.
- Using the unequal two-toss model, find the probability of a head on the first toss and on the second toss. Why need these probabilities not be equal in an arbitrary assigned model?
Key Takeaways
• Probability assigns weights to events using nonnegativity, total weight one, and disjoint additivity. • The axioms imply P(∅)=0 and probabilities between zero and one. • Valid outcome assignments require nonnegative entries whose total is exactly one. • P(A) is the sum of the weights of the outcomes in A. • Counting labels alone does not establish equal probabilities.