a Set of N distinct objects can be arranged in

N! Permutations, or orderings N items, pick 1; N-1 items remain

N-1 items, pick 1; N-2 items remain

N-2 items, pick 1; N-3 items remain...

1 item left, only choice to pick

N (N-1) (N-2) ... 1 possible selected

N! possible Permutations Permutations of k < n objects Tossing a single coin, n times Singular outcome:

a particular sequence of heads and tails

{ k successive heads in any order } number of elementary events

of k heads, n-k tails in n tosses probability of each elementary event P{ 0 heads in 4 tosses } = 1 * (1/2)^0 * (1/2)^4 = .063

P{ 1 head in 4 tosses } = 4* (1/2)^1 * (1/2)^3 = .250

P{ 2 heads in 4 tosses } = 6 * (1/2)^2 * (1/2)^2 = .375

P{ 3 heads in 4 tosses } = 4 * (1/2)^3 * (1/2)^1 = .250

P{ 4 heads in 4 tosses } = 1 * (1/2)^4 * (1/2)^0 = .063 Combinations of k < n objects

(taken without respect to order) Why? Only taking k objects, not all n Factor out the k! permutations for each group of k objects which combine to a single combination Note: we now have a formula

for the total number of k-object subsets of a set of n objects Now, consider an Experiment n=20

k=12 n-k=8 The likelyhood of k successive heads in n tosses, then, is

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# Bernoulli Trials

Example 3-6 from Probability, Random Variables, and Stochastic Processes, 4th ed. Athanasios Papoulis

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