Probabilità/Calcolo combinatorio: differenze tra le versioni

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== Combinazioni ==
 
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Permutations and Combinations[edit]
Permutations
arrangement of objects without repetition where order is important. Permutations using all the objects: n objects, arranged into group size of n without repetition, and order being important – P(n, n) = N!
Example: Find all permutations of A, B, C.
Permutations of some of the objects: n objects, group size r, order is important. P(n, r) = N!/(n-r)! Example: Find all 2-letter combinations using letters A, B, C.
 
Distinguished permutations
if a word has n letters, k of which are unique, let n (n1, n2, n3....nk) be the frequency of each of the k letters. N!/(n1!)(n2!)(n3!)
Combinations
arrangement of objects without repetition, where order is NOT important. A combination of n objects, arranged in groups of size r, without repetition, and order being important. C(n, r) = N!/r!(n-r)!
Another way to write a combination of n things, r at a time, is using the Binomial notation (Binomial Distribution), sometimes described as "n choose r".
 
Counting Rules[edit]
Rule 1
If any one of K mutually exclusive and exhaustive events can occur on each of N trials, there are KN different sequences that may result from a set of such trials
Example: Flip a coin three times, finding the number of possible sequences. N=3, K=2, therefore, KN =(2)(3)=6
Rule 2
If K1, K2, ....KN are the numbers of distinct events that can occur on trials 1,....N in a series, the number of different sequences of N events that can occur is (K1)(K2)...(KN)
Example: Flip a coin and roll a die, finding the number of possible sequences. Therefore, (K1)(K2) = (2)(6) = 12
Rule 3
The number of different ways that N distinct things may be arranged in order is N! = (1)(2)(3)....(N-1)(N), where 0! = 1. An arrangement in order is called a permutation, so that the total number of permutations of N objects is N! (the symbol N! Is called N-factorial)
Example: Arrange 10 items in order, finding the number of possible ways. Therefore, 10! = 10x9x8x7x6x5x4x3x2x1 = 3628800
Rule 4
The number of ways of selecting and arranging r objects from among N distinct obejects is: N!/(N-r)! [nPr]
Example: pick 3 things from 10 items, and arrange them in order. Therefore N=10, r=3, so 10!/(10-3)! = 10!/7! = 720
Rule 5
The total number of ways of selecting r distinct combinations of N objects, irrespective of order (ie order NOT important), is: N!/r!(N-r)! [nCr]
Example: Pick 3 items from 10 in any order, where N=10, r=3. Therefore, 10!/3!(7!) = 720/6 = 120
Consequences[edit]
We can now give some basic theorems using our axiomatic probability space.
 
Theorem 1[edit]
Given a probability space (Ω,S,P), for events A,B\in S:
 
P(A\cup B)=P(A)+P(B)-P(A\cap B)
 
 
Axioms of Probability[edit]
A probability function has to satisfy the following three basic axioms or constraints: To each event a a measure (number) P(a) which is called the probability of event a is assigned. P(a) is subjected to the following three axioms:
 
P(a) ≥ 0
P(S) = 1
If a∩b = 0 , then P(a∪b) = P(a)+ P(b)
Corollaries
P(0) = 0
P(a) = 1- P(a)≤ 1
If a ∩b ≠ 0 , then P(a ∪ b) = P(a)+ P(b) - P(a ∩ b)
If b ⊂ a , P(a) = P(b)+ P(a ∩ b)≥ P(b)
 
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== Partizioni ==