A probability space separates outcomes, events, and assigned probabilities
Objective: Why is this not the ordinary fair-die answer by default?
One die roll produces exactly one outcome such as 4. An event is a question represented by a group of outcomes, such as “the result is even,” represented by {2,4,6}. A probability model assigns one number from zero to one to each possible outcome; for discrete outcomes that number is formally called probability mass, but it means probability, not physical weight.
What must be established: For every event A, 0≤P(A)≤1, P(∅)=0, P(Ω)=1, and P(Aᶜ)=1−P(A). First use the definition “For a finite model, Ω is the complete outcome list and A is any selected group inside it. The notation p(ω) is the probability of one outcome ω. All individual probabilities must be nonnegative and add to one. The event probability P(A) is found by adding p(ω) only for outcomes belonging to A.” The first nearby case where the conclusion can fail is: Listing only “success” and “failure” is invalid when the experiment’s complete outcomes or their probabilities are not defined.
Read the notation before calculating: Read Ω as “the list of all possible results.” A is one chosen group of results. The symbol ω means one particular result inside Ω. P(A) means “the probability that the result belongs to A.” The notation p(ω) means “the probability assigned to that one result.” Σ means “add the following quantity once for every listed result”; it is an instruction to add, not a new unknown.
For every event A, 0≤P(A)≤1, P(∅)=0, P(Ω)=1, and P(Aᶜ)=1−P(A).
Every single-outcome probability is at least zero, so adding probabilities for an event cannot produce a negative result.
The event uses only some outcomes from Ω, while all outcome probabilities together equal one; therefore the event probability cannot exceed one.
A and its complement Aᶜ contain every outcome exactly once between them, so P(A)+P(Aᶜ)=1.
A loaded six-sided die is modeled as follows: faces 1, 2, 3, and 4 each have probability 1/8; faces 5 and 6 each have probability 2/8=1/4. Find the probability of rolling an even number.
- The complete outcome list is Ω={1,2,3,4,5,6}. The event “even” contains exactly the outcomes {2,4,6}.
- Read the six assigned probabilities in face order: P(1)=1/8, P(2)=1/8, P(3)=1/8, P(4)=1/8, P(5)=2/8, and P(6)=2/8. Therefore the required probabilities are P(2), P(4), and P(6).
- Add 1/8+1/8+2/8=4/8=1/2. As a check, the odd outcomes have probability 1/8+1/8+2/8=1/2, and the two complementary probabilities add to one.
Result: P(even)=1/2.