STEP-BY-STEP LESSON · §9.8
Probability Axioms and Expected Value
Compute an average weighted by probabilities.
Before you begin
Fractions
a/b is a quotient with b≠0. For integer a,b it represents a rational number. When adding fractions, put them over a common nonzero denominator; keep exact values until the final result.
A finite probability model
Specify the finite set of possible outcomes and their probabilities. Probabilities are nonnegative and total 1. Favorable count divided by total count is valid only when those elementary outcomes are equally likely.
Symbols
- E(X)
- expected value of X
- Σx·P(X=x)
- Sum over all distinct values x of the finite random variable X
Definitions and notation for this topic
Probability
P(A)
A numerical measure of how likely an event is, between 0 and 1, assigned by the probability model.
Counting favorable outcomes and dividing by the total requires finitely many equally likely outcomes.
The formula P(A)=|A|/|S| requires a finite sample space with equally likely outcomes; it is not a general definition for every model.
Separate glossary example
For a fair die, P({2,4,6})=3/6.
A finite probability model
P(A)=∑_{ω∈A}p_ω; ∑_{ω∈S}p_ω=1On a finite nonempty sample space S, assign each outcome ω a weight p_ω≥0 with total 1. The probability of event A⊆S is the sum of its outcome weights.
This gives P(∅)=0, P(S)=1 and additivity for disjoint events. The counting shortcut |A|/|S| is valid only when every outcome has weight 1/|S|.
Separate glossary example
If S={a,b,c} has weights 1/2,1/3,1/6, then P({a,b})=5/6, not 2/3.
Random variable
X:S→ℝ; {X=x}A random variable assigns a real number X(ω) to each outcome ω. The notation {X=x} abbreviates the event {ω∈S:X(ω)=x}.
Different outcomes can give the same value. P(X=1) sums the probabilities of HT and TH; X itself is not an event.
Separate glossary example
For two coin tosses, let X count heads: X(TT)=0, X(HT)=X(TH)=1, X(HH)=2.
Expected value: finite case
E(X)=∑_x xP(X=x)
For a random variable with finitely many possible real values, multiply each distinct value by its probability and sum over those values.
The expectation need not be an attainable value. The sum is over values of X, not over repeated copies of the same value; infinite distributions need additional convergence conditions not covered by this finite formula.
Separate glossary example
For a fair six-sided die whose result is X, E(X)=(1+2+3+4+5+6)/6=3.5.
Step by step
Step 1 / 6
Assign probabilities
Probabilities are nonnegative, the whole space has probability 1, and probabilities of disjoint events add.
These rules constrain a valid model.
Worked example
A game has net outcome +4 with probability 1/4 and −2 with probability 3/4. Find its expectation.
- The two possible net outcomes exhaust the game: 1/4+3/4=1. They already include any fee.
- Winning contribution: 4·1/4=1; losing contribution: −2·3/4=−1.5.
- E(X)=1−1.5=−0.5. This is an average, not the result of one play.
Optional self-check
A game costs 2 and pays 6 with probability 1/4, otherwise 0. What is expected net profit, and can it be a single outcome?
Show answer
Expected payout is 6/4=1.5; expected net is −0.5. Single net outcomes are 4 and −2, so −0.5 is not a possible single outcome.