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Unit 10 · 2.6

Probability with combinations

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

Practice 0/15Checkpoint best 0%

Rule cards

Probability = ways you want ÷ all ways

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

  1. Total ways: C(total, picked). TVs: C(10,3) = 120.
  2. Wanted ways: one C per pool, multiplied.
  3. Check: all cases add to the total (8 + 56 + 56 = 120).
  4. Identical items (socks) still count as separate objects: red pair 55/171.
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

At least one = 1 − none

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

Dice sums: list the combos, then count the orders

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

  1. Washers in size order: 2 good orders ÷ 5! = 2/120 = 1/60.
  2. 2 prizes, 50 tickets, you hold 2: C(2,2)/C(50,2) = 1/1225.
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

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