Discrete Mathematics University of Kentucky CS 275 Spring ... - MGNet

Theorem: Let E and F be events in a sample space S. Then
p(E2F) = p(E) + p(F) – p(E3F).
Proof: Recall that |E2F| = |E| + |F| – |E3F|. Hence,
p(E2F) = |E2F| / |S| = (|E| + |F| – |E3F|) / |S| = p(E) + p(F) – p(E3F).
Example: What is the probability in the set {1, 2, …, 100} of an element being
divisible by 2 or 3? Let E and F represent elements divisible by 2 and 3,
respectively. Then |E| = 50, |F| = 33, and |E3F| = 16. Hence, p(E2F) = 0.67.
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Second Assumption: Now suppose that the probability of an event is not 1 / |S|.
In this case we must assign probabilities for each possible event, either by
setting a specific value or defining a function.
Definition: For a sample space S with a finite or countable number of events, we
assign probabilities p(s) to each event s,S such that
(1) 0 ; p(s) ; 1 (s,S, and
(2) " p(s) = 1.
s!S
Notes:
1. When |S| = n, the formulas (1) and (2) can be rewritten using n.
2. When |S| = . and is uncountable, integral calculus is required for (2).
3. When |S| = . and is countable, the sum in (2) is true in the limit.
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