Unit 10 · 2.6

Probability with combinations

C(pool, want) × … ÷ C(total, picked); at least one = 1 − none

Locked

Pass Unit 9 (Permutations vs combinations) with 4 of 5 on its checkpoint to unlock this unit. Exam is tomorrow, so you can also open it now.

Rule cards

P = C(pool₁, want₁) × C(pool₂, want₂) ÷ C(total, picked); the 'want' numbers add up to the number picked.

Traps

Used permutations on top and combinations on the bottom.
P(2 bad) → P(2,2)·P(8,1) ÷ C(10,3)
✓ use C everywhere: C(2,2)·C(8,1) ÷ C(10,3) = 8/120 = .067 (exam Q25)
Counted only the pool you care about.
exactly 2 bad → C(2,2) ÷ C(10,3)
✓ the 3rd TV must be good: × C(8,1)
Took the counts from the wrong pool.
2 men 1 woman → C(women,2)·C(men,1)
✓ match each 'want' to its own pool

P(at least one) = 1 − P(none); P(at least two) = 1 − P(0) − P(1).

Traps

Reported P(none) when 'at least one' was asked.
orchestras → .083
✓ 1 − .083 = .917
Rounded too early before subtracting.
1 − .08 − .32 = .60
✓ keep 3+ decimals: 1 − .083 − .323 = .594

For 3 dice, list each set of faces that works, then count its orders: 3 different faces → 6, one pair → 3, all same → 1.

Traps

Counted each combo once.
sum 7 → 4 combos → 4/216
✓ 1-1-5 (3) + 1-2-4 (6) + 1-3-3 (3) + 2-2-3 (3) = 15 → 15/216
Divided top and bottom by different numbers.
15/216 → 3/72
✓ ÷3 both: 5/72 (the key's typo — see Key errors)
Used the 2-dice sample space.
sum 7 → 6/36
✓ 3 dice → 216 outcomes