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2.2.2 Combinations
Definició 20 The number of different combinations of n objects in groups of r is the number of different non-ordered groups that can be constructed with r objects taken from a set with n objects
(i) Without replacement
An easy way to find the number of combinations is to notice that for each of these combinations, that is, for each of these groups with r objects without order we can form r · (r − 1 ) · (r − 2 ) · · · ( 2 ) · ( 1 ) = P(r, r) = r! ordered lists.
Thus, the total number of ordered lists that we can produce with r objects taken from a set with n objects , P(n, r), is equal to the number of different combinations, C(n, r), multiplied by the number of ordered lists we can form with each of these combinations, P(r, r).
That is, P(n, r) = C(n, r) · P(r, r)
or, using the formula to compute permutations,
n! (n − r)!
= C(n, r) · r!
Therefore , to count the number of combinations we have the formula
C(n, r) =
n r
n! r! · (n − r)!
Exemple 22 In a grocery store there are 10 types of fruits. We want to buy 3 kilos of fruit, how many combinations can we have?
C( 10 , 3 ) =
2.3 Probability: axiomatic definition and interpretations
- We seek now a function (or formula) that assigns to each event the proba- bility that it occurs.
- The modern approach to probability is based on the “axiomatic definition” by Kolmogorov, that we will see next
- The classical (by Laplace) and frequentialist approaches can be understood as particular applications of interpretations of that broader approach.
2.3.2 Probability interpretations
(a) The classical interpretation of probability by Laplace claims that the prob- ability that the event A occurs is the ratio between the number of outcomes favorable to the event (the cardinality of the set A, denoted nA) divided by the total number of possible outcomes (the cardinality of the sample space Ω, denoted nΩ) P(A) =
nA nΩ
For instance, that probability that we get “Heads” when tossing a coin is 0. because there is one outcome favorable and two possible outcomes.
It is easy to verify that this approach to probability satisfies Kolmogorov’s 3 axioms (provided the sample space is not empty!):
(i) Since whatever the event A is we will have nA ≥ 0 and nΩ > 0 , then it is clear that P(A) ≥ 0
(ii) P(Ω) = n nΩΩ = 1
(iii) If A ∩ B = 0 we will have that/ nA∪B = nA + nB. Hence,
P(A ∪ B) =
nA∪B nΩ
nA + nB nΩ
nA nΩ
nB nΩ
= P(A) + P(B)
This probability interpretation also satisfies the three Kolmogorov axioms:
(i) Since for whatever event A we have that fA(t) ≥ 0, the limit will also be non-negative
(ii) Clearly, fΩ = 1
(iii) If A ∩ B = 0 we will have/ fA∪B = fA + fB.
This probability interpretation also has problems when trying to compute the probability of an event. For instance, if we want to find the probability of Heads when tossing a coin,
- In order to conclude that the probability of Heads is (exactly) 12 , how many times do we need to toss the coin to reach this conclusion?
- How can we be sure that each toss of the coin is done under the same initial conditions?
- How can we be sure that the coin is being tossed really at random?
