🎲 Sessions 2.1–2.3 Bridge: Sets → Counting → Probability

When outcomes are equally likely, P(event) = |favourable| / |total|. Section 2.2 teaches us to count those sets; Section 2.3 turns those counts into probabilities.

Bridge: 2.2 → 2.3
P(event) = |event outcomes| ÷ |all outcomes|

Sets tell us which outcomes to count (OR/AND/NOT ⇔ union/intersection/complement).
Counting (permutations, combinations) tells us how many outcomes there are.

① OR → Union → Inclusion–Exclusion

To count A OR B, use |A ∪ B| = |A| + |B| − |A ∩ B|, then divide by total.

Example: Heart OR King from a deck.
|Hearts| = 13, |Kings| = 4, |Heart ∩ King| = 1 (King of Hearts).
|A ∪ B| = 13 + 4 − 1 = 16; Total = 52 → P(Heart ∪ King) = 16/52 = 4/13.
Add the sizes, subtract the overlap (to avoid double-counting).
Convert count → probability by dividing by total outcomes.

② Exactly k successes (Sampling without replacement)

When drawing without replacement from two groups, the count of favourable samples is a product of combinations.

Example: 50 parts (3 defective). Draw 6. P(exactly 2 defective).

Favourable count = C(3, 2) · C(47, 4)
Total count = C(50, 6)
⇒ P(X = 2) = [C(3,2) · C(47,4)] / C(50,6)
C(3,2) = 3!/(2!·1!) = 3   and   C(47,4) = (47·46·45·44)/(4·3·2·1) = 178,365
C(50,6) = 50!/(6!·44!) (leave as a count).
Probability = (3 · 178,365) / C(50,6).

This is the hypergeometric model: combinations from 2.2 directly power 2.3 probabilities.

③ “At least one” via Complement

Often it’s faster to count the complement (none) and subtract from 1.

Same setting: 50 parts, 3 defective, draw 6.
P(≥ 1 defective) = 1 − P(none defective) = 1 − C(47,6) / C(50,6).
Counting “at least one” requires multiple cases (1, 2, …). Counting “none” is a single combination.

④ Permutation vs Combination (Order?)

Ask: Does swapping positions create a new outcome?

If yes → Permutation: P(n,r) = n!/(n−r)!
If no → Combination: C(n,r) = n!/[r!(n−r)!]
P(8,4) = 8×7×6×5 = 1,680 (order matters)
C(8,4) = (8×7×6×5)/(4×3×2×1) = 70 (order doesn’t)

In uniform spaces, probability of one specific ordered 4-tuple is 1/P(8,4); of one specific 4-set is 1/C(8,4).

🧭 Takeaway for 2.3