Probability Rules: Addition & Multiplication
Probability has a few core rules that let you combine events. Want to know the chance of A or B happening? That’s the addition rule. Want A and B? That’s the multiplication rule. These two rules, once mastered, let you solve almost any probability problem.
Part 1: The Addition Rule
The probability of A or B (or both) happening is:
Why subtract P(A and B)? Because when we add P(A) and P(B), we double-count the overlap — outcomes where both happen. We need to subtract it once to fix that.
Visualize A and B as two overlapping curves. The total “or” probability is everything under either curve, but the overlap region shouldn’t be counted twice:
Try setting the overlap to 0. When P(A and B) = 0, the events are mutually exclusive — they can’t both happen. The formula simplifies to: P(A or B) = P(A) + P(B). Notice how the curves don’t overlap at all!
Part 2: Mutually Exclusive Events
When two events cannot happen at the same time (like rolling a 2 and a 5 on one die), they are mutually exclusive. Their overlap is zero:
As you increase the separation, the curves move apart and the overlap vanishes. Completely separated = mutually exclusive. For these events, simple addition works:
Part 3: The Multiplication Rule
For independent events, the probability of both happening is:
Think of a tree diagram: first A happens (or doesn’t), then B happens (or doesn’t). Each branch multiplies the probabilities along its path.
Let’s model two independent events with adjustable probabilities:
Tree diagram thinking: Imagine flipping two coins.
- P(heads on first) = 0.5
- P(heads on second) = 0.5
- P(both heads) = 0.5 * 0.5 = 0.25
Each path through the “tree” multiplies the branch probabilities. Set both sliders to 0.5 and verify the joint probability shows 0.25!
Part 4: Dependent Multiplication Rule
When events are not independent, the multiplication rule uses conditional probability:
The probability of B changes depending on whether A happened:
Card example: Draw 2 cards from a deck without replacement.
- P(first is Ace) = 4/52
- P(second is Ace | first was Ace) = 3/51 (one ace is gone!)
- P(both Aces) = (4/52) * (3/51) = 12/2652
Notice how P(B|A) is smaller than P(B) would be alone — the first draw removed an ace from the deck. That’s dependence in action.
Part 5: Complement Rule
Sometimes it’s easier to calculate what you don’t want and subtract:
Challenge: You roll a die 3 times. What’s the probability of getting at least one 6?
Hint: It’s easier to find P(no sixes at all) and subtract from 1. P(not 6 on one roll) = 5/6. For three independent rolls: P(no sixes) = (5/6)^3. So P(at least one 6) = 1 - (5/6)^3.
Calculate it, then verify: is it higher or lower than you’d guess?
Wrapping Up
| Rule | Formula | When to Use |
|---|---|---|
| Addition | P(A or B) = P(A) + P(B) - P(A and B) | Either event happening |
| Mutually Exclusive | P(A or B) = P(A) + P(B) | Events can’t co-occur |
| Multiplication (independent) | P(A and B) = P(A) * P(B) | Both events, no influence |
| Multiplication (dependent) | P(A and B) = P(A) * P(B|A) | Both events, with influence |
| Complement | P(A) = 1 - P(not A) | “At least one” problems |
These five rules are your complete toolkit. Every probability problem you’ll encounter in this course can be broken down into combinations of these rules.