Definition D1
For an event A of a sample space S, the event Aᶜ = S ∖ A is called the complement of A.
Proof.
(1)Every outcome of S belongs to A or to Aᶜ, so S = A ∪ Aᶜconvention
(2)No outcome belongs to both, so A ∩ Aᶜ = ∅convention
(3)Therefore A and Aᶜ are disjoint and form a partition of S.(1), (2)
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Definition D2
When S is finite and all outcomes are equally likely, the probability of an event E is defined as P(E) = |E| / |S|.
Proof.
This definition applies only to finite sample spaces with equally likely outcomes.convention
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Theorem T1
For every event A, P(Aᶜ) = 1 − P(A)
Proof.
(1)S = A ∪ Aᶜ, and the two events are disjoint.D1
(2)P(S) = P(A) + P(Aᶜ)(1), A2
(3)P(S) = 1A1
(4)P(A) + P(Aᶜ) = 1(2), (3), E
(5)P(Aᶜ) = 1 − P(A)(4), A3
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Corollary C1
When a die is rolled n times, the probability that a 6 appears at least once is 1 − 5ⁿ/6ⁿ.
Proof.
(1)Suppose a single roll has 6 outcomes, all equally likely.assumption
(2)The n rolls have 6 options in each position, so |S| = 6ⁿ(1), A4
(3)Let A be the event that a 6 appears at least once. Then Aᶜ is the event that a 6 never appears.D1
(4)The non-6 outcomes give 5 options in each position, so |Aᶜ| = 5ⁿ(3), A4
(5)P(Aᶜ) = 5ⁿ / 6ⁿ(1), (2), (4), D2
(6)P(A) = 1 − 5ⁿ/6ⁿ(5), T1
(7)To count A directly, one must split it into the events where a 6 appears exactly k times. Those events are mutually disjoint, so |A| is a sum of n terms.(3), A5
(8)Aᶜ does not split. One application of the multiplication principle — 5 options in each position — finishes it.(4), A4
(9)For n = 2, P(A) = 1 − 25/36 = 11/36.(6), N
(10)Counting directly: a 6 appears exactly once in 2 × 5 = 10 ways and exactly twice in 1 way, so |A| = 11 and the probability is 11/36. The two values agree.(7), (9), A4, A5, D2, N
(11)The complement is easy not because the computation is short. The opposite of 'at least one' is 'none at all', and that is a single event with the same condition in every position.(7), (8)
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