Probability as favorable outcomes over total outcomes. Coin flips, dice rolls.
Deep-dive lesson - accessible entry point but dense material. Use worked examples and spaced repetition.
Mini‑puzzle (predict before reading):
You roll two fair six‑sided dice and add them.
Question: Which sum is more likely: 6 or 7?
Don’t compute yet—just predict.
Now open the interactive canvas for this node:
Hold onto your prediction. By the end, you’ll be able to justify it using just one idea: probability = favorable outcomes / total outcomes (when outcomes are equally likely).
Basic probability (classical probability) starts with a sample space Ω (all possible outcomes) and an event A ⊂ Ω (outcomes you care about). If all outcomes in Ω are equally likely, then $$
In many everyday situations you can’t predict the exact outcome, but you can still reason about how likely different outcomes are.
Probability gives a precise language for “chance.” In this node we focus on the simplest, most concrete version: classical probability.
1) Sample space (usually written Ω)
2) Event (often written A)
We write P(A) for “the probability that event A happens.”
In this node, we restrict to situations where outcomes are equally likely (fair coin, fair die, well‑shuffled cards, etc.). In that setting, the probability of an event is:
Use the canvas as a counting tool:
This is not just a visualization—it’s the workflow you’ll use repeatedly.
A very common beginner mistake is to compute “favorable / total” with the wrong “total.” The total is not “whatever feels like all outcomes”—it’s exactly the size of your sample space Ω.
So we slow down and build a habit:
1) Name the experiment (what is being done?)
2) Write Ω (all outcomes)
3) Describe A as a subset of Ω
4) Count |Ω| and |A|
Experiment: flip a fair coin once.
Experiment: roll a fair six‑sided die.
When you repeat an experiment, each outcome often becomes a tuple.
Experiment: flip a coin twice.
A good sample space is ordered outcomes:
Event A = “exactly one head”
Notice how order matters: HT ≠ TH as outcomes (even if they both represent “one head”).
In the interactive canvas, two dice outcomes are naturally shown as a 6×6 grid.
So if you can count highlighted cells, you can compute probability.
Because events are sets, you can combine them.
These lead to quick probability facts in the equally‑likely setting:
And if A and B do not overlap (A ∩ B = ∅), then
You don’t need more set theory than that for this node—just recognize events can be built and combined.
In classical probability, the hardest part is usually not “probability” itself—it’s counting.
You already know counting principles (addition and multiplication rules). Here’s how they plug into probability:
When outcomes are equally likely:
The word “equally likely” is doing a lot of work.
If outcomes are not equally likely, you need more advanced tools (later nodes). For now, we deliberately stay in the equally‑likely world.
Experiment: roll two dice and sum them.
Step 1: sample space
Step 2: event for sum = 6
A₆ = {(d₁,d₂) : d₁+d₂=6}
List them:
So |A₆| = 5 and P(sum=6)=5/36.
Step 3: event for sum = 7
A₇ = {(d₁,d₂) : d₁+d₂=7}
List them:
So |A₇| = 6 and P(sum=7)=6/36=1/6.
Conclusion: 7 is more likely than 6 because there are more ordered pairs that add to 7.
On the interactive canvas:
1) Toggle to “Two dice (grid)” mode.
2) Click “Highlight sum = 7.” Count highlighted cells: 6.
3) Click “Highlight sum = 6.” Count highlighted cells: 5.
4) The fraction highlighted/36 is the probability.
This is the exact same favorable/total rule—just visual.
| Experiment | Sample space Ω | Ω | ||
|---|---|---|---|---|
| 1 fair coin flip | {H,T} | 2 | ||
| 2 fair coin flips | {HH,HT,TH,TT} | 4 | ||
| 1 fair die roll | {1,2,3,4,5,6} | 6 | ||
| 2 fair dice | {(d₁,d₂)} | 36 |
If a die is loaded, or a coin is biased, you cannot assume P(face) = 1/6.
Classical probability is best viewed as:
Later, you’ll generalize to probability models where outcomes have different weights.
A random variable is a rule that assigns a number to each outcome in Ω.
In the two‑dice example:
X is a random variable: it turns outcomes into numbers.
Why this matters:
Then probability becomes:
That is the same favorable/total idea, just written in a way that scales.
Conditional probability asks: how does the probability change once you learn something?
Even in the equally‑likely world, learning information shrinks your sample space.
Example (two dice):
Now ask: what is the probability the sum is 7 given die 1 is 6?
Event A = “sum is 7.” Within Ω′, only (6,1) works.
This is the intuition behind :
You will formalize this in the Conditional Probability node.
Whenever you feel stuck, write:
If you can answer those, you can compute P(A) in this basic setting.
A standard deck has 52 cards, well shuffled. Find P(A) where A = “the card is a heart.” Assume each card is equally likely.
Define the sample space: Ω = set of all 52 distinct cards in the deck.
So |Ω| = 52.
Define the event: A = {all hearts}.
In a standard deck there are 13 hearts, so |A| = 13.
Apply classical probability:
Simplify the fraction:
Insight: When outcomes are equally likely, probability is just a fraction of the sample space. The hard part is knowing what’s in Ω and counting correctly.
Flip a fair coin twice. Find the probability of the event A = “at least one head.”
Write the sample space of ordered outcomes:
Ω = {HH, HT, TH, TT}.
So |Ω| = 4.
Describe the event A:
“At least one head” includes HH, HT, TH.
So A = {HH, HT, TH} and |A| = 3.
Compute:
Optional cross-check using complement:
Let B = “no heads” = {TT}.
Then P(B) = 1/4, so
Insight: The complement trick (1 − P(complement)) is often easier when the event is phrased as “at least one…” or “not…”
Roll two fair six-sided dice. Let A be the event “sum ≥ 10.” Find P(A).
Define the sample space:
Ω = {(d₁,d₂) : d₁,d₂ ∈ {1,…,6}}.
So |Ω| = 6×6 = 36.
List favorable outcomes by sums.
Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes.
Sum = 11: (5,6), (6,5) → 2 outcomes.
Sum = 12: (6,6) → 1 outcome.
Count favorable outcomes:
|A| = 3 + 2 + 1 = 6.
Compute:
Insight: For two dice, thinking in ordered pairs prevents undercounting. The canvas grid makes “36 equally likely outcomes” feel concrete.
A sample space Ω is the set of all possible outcomes of an experiment.
An event A is a subset of the sample space: A ⊂ Ω.
In classical probability (equally likely outcomes), $$.
For multi-step experiments, outcomes are often ordered tuples (like (d₁,d₂) for two dice).
The complement rule is a fast tool: P(Aᶜ) = 1 − P(A).
Union/intersection correspond to “or/and” combinations of events; disjoint unions add probabilities.
The interactive canvas is a counting aid: highlight the event inside Ω, then compute highlighted/total.
Using the wrong sample space (wrong “total outcomes”), especially when order matters (HT vs TH).
Assuming outcomes are equally likely when they aren’t (loaded die, biased coin, non-uniform mechanism).
Counting favorable outcomes correctly but forgetting to divide by |Ω| (or dividing by the wrong number).
Mixing up “at least one” with “exactly one,” or failing to use the complement when it would simplify.
Roll a fair six-sided die once. What is the probability of rolling a number greater than 2?
Hint: Write Ω = {1,2,3,4,5,6}. Your event is {3,4,5,6}.
Ω has 6 outcomes. Event A = {3,4,5,6} has 4 outcomes.
Flip a fair coin three times. What is the probability of getting exactly two heads?
Hint: There are 2³ outcomes total. Count outcomes with exactly two H’s (think of positions for the tails).
Sample space size: |Ω| = 2³ = 8.
Exactly two heads means exactly one tail. Choose which flip is T:
So |A| = 3.
Roll two fair dice. What is the probability that at least one die shows a 6?
Hint: Use the complement: 1 − P(no 6’s). No 6 on one die has probability 5/6, so for two dice it’s (5/6)².
Let A = “at least one 6.” Use complement Aᶜ = “no 6’s.”
For one die: P(no 6) = 5/6.
For two independent dice, the sample space is 36 equally likely pairs, and:
So
Unlocks and next steps:
Related prior knowledge: