Counting and probability

When the sample space is huge, nobody lists it. Count it with permutations and combinations instead, and count the favourable outcomes the same way.

When order does not matter, use combinations. Drawing cards, choosing balls or picking a committee: only which ones come out matters.

• Two cards from 52: there are ⁵²C₂ = 1326 possible pairs. P(both aces) = ⁴C₂/⁵²C₂ = 6/1326 = 1/221.
• Three cards, all the same colour: (²⁶C₃ + ²⁶C₃)/⁵²C₃ = 5200/22100 = 4/17.

Balls in a bag: 5 red, 4 green, 3 blue (12 in all); two are drawn. Total: ¹²C₂ = 66.

• Both red: ⁵C₂/66 = 10/66 = 5/33.
• One red and one green: (⁵C₁ × ⁴C₁)/66 = 20/66 = 10/33. Multiply when the choices happen together.
• Both the same colour: (⁵C₂ + ⁴C₂ + ³C₂)/66 = (10 + 6 + 3)/66 = 19/66. Add when the cases are separate.

'At least one' is usually easiest as 1 – P(none). Three balls from the same bag: P(at least one blue) = 1 – ⁹C₃/¹²C₃ = 1 – 84/220 = 34/55.

Good and faulty items. 12 bulbs, 4 of them faulty; 3 are picked.

• Exactly one faulty: (⁴C₁ × ⁸C₂)/¹²C₃ = (4 × 28)/220 = 28/55.
• None faulty: ⁸C₃/220 = 56/220 = 14/55.

When order matters, use permutations or factorials. Arrangements, finishing positions and codes:

• The letters of WORD in random order: 4! = 24 arrangements. P(it spells WORD) = 1/24. P(W comes first) = 3!/4! = 1/4.
• P(W just before O): glue WO into one block, giving 3! = 6 arrangements, so 6/24 = 1/4.
• A 3-digit code with no digit repeated: 10 × 9 × 8 = 720 codes, so one guess opens it with probability 1/720.

Three events. The addition rule extends: P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(A ∩ C) – P(B ∩ C) + P(A ∩ B ∩ C). With 0.5, 0.4, 0.3, overlaps 0.2, 0.1, 0.15 and 0.05 for all three, the total is 0.8.

Read the conditions exactly. Four-digit numbers made from 1, 2, 3, 4 without repetition: 24 in all. Those greater than 3000 must start with 3 or 4, giving 2 × 3! = 12, so the probability is 1/2. Words like 'greater than', 'at least' and 'exactly' each change what you count.