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Lectures on Elementary Probability

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2.5. BINOMIAL PROBABILITIES 11<br />

Here #E is the number of elements in the set E, while #S is the number of<br />

points in the set S. The uniform probabilities satisfy all the probability axioms.<br />

We shall see in the next secti<strong>on</strong>s that this noti<strong>on</strong> has a number of important<br />

and useful special cases.<br />

• The sample space c<strong>on</strong>sists of all functi<strong>on</strong>s from a finite set to another finite<br />

set.<br />

• The sample space c<strong>on</strong>sists of all injective functi<strong>on</strong>s from a finite set to<br />

another finite set.<br />

• The sample space c<strong>on</strong>sists of all subsets of a finite set, such that each<br />

subset has a certain specified number of elements. (This is the same as<br />

all functi<strong>on</strong>s from the set to a two-point set, such that each functi<strong>on</strong> has<br />

certain specified occupati<strong>on</strong> numbers.)<br />

• The sample space c<strong>on</strong>sists of all functi<strong>on</strong>s from a finite set to another finite<br />

set, such that each functi<strong>on</strong> has certain specified occupati<strong>on</strong> numbers.<br />

2.5 Binomial probabilities<br />

In this secti<strong>on</strong> we c<strong>on</strong>sider an n element set I of trials and an m element set M<br />

representing a populati<strong>on</strong>. The fundamental assumpti<strong>on</strong> is that every ordered<br />

sample with replacement has the same probability. In other words, this is the<br />

assumpti<strong>on</strong> of uniform probability <strong>on</strong> the set of all functi<strong>on</strong>s from I to M.<br />

The number of ordered samples with replacement is m n . Therefore each<br />

ordered sample with replacement has probability<br />

P [f] = 1<br />

m n . (2.8)<br />

Let A be a subset of M representing the successes. Suppose that A has a<br />

elements.<br />

Theorem 2.1 The probability that an ordered sample with replacement has exactly<br />

k successes is<br />

where p = a/m.<br />

( n a<br />

P (k) =<br />

k) k (m − a) n−k ( ) n<br />

m n = p k (1 − p) n−k . (2.9)<br />

k<br />

Proof: Let K be a subset of the set of trials I with k elements. The probability<br />

P (K) that an ordered sample with replacement has successes that exactly<br />

bel<strong>on</strong>g to the set K is the number of such samples time the probability of each<br />

sample. The number of such samples is the number of functi<strong>on</strong>s from K to A

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