By Henri Poincaré
Excerpt from Calcul des Probabilités
Le most suitable exemple que nous allons choisir est celui de l'équilibre instable; si un cône repose sur sa pointe, nous savons bien qu'il va tomber, mais nous ne savons pas de quel côté; il nous semble que le hasard seul va en décider. Si le cône était parfaitement symétrique, si son awl était_ parfaitement vertical, s'il n'était soumis à aucune autre strength que los angeles pesanteur, il ne tomberait pas du tout. Mais le moindre défaut de symétrie va le faire pencher légèrement d'un côté ou de l'autre, et dès qu'il penchera, si peu que ce soit, il tombera tout à fait de ce côté. Si même los angeles symétrie est parfaite, une trépidation très légère, un soufi°le d'air pourra le faire incliner dequelques secondes d'arc; ce sera assez pour déterminer sa chute et même le sens de sa chute qui sera celui de l'inclinaison initiale.
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Additional info for Calcul des probabilites
Example text
An−1 ) > 0 anche P(A1 ∩ A2 ∩ . . 35) sono ben definite. La dimostrazione procede per induzione. 35) è verificata per n = 2. 35). 2 Bayes e dintorni In molte situazioni, la nozione di probabilità condizionale è utile nella costruzione stessa di un modello probabilistico: talvolta è “naturale” assegnare il valore di alcune probabilità condizionali, e da esse dedurre il valore di probabilità non condizionali. 46. Due urne, che indichiamo con α e β , contengono rispettivamente 3 palline rosse e 1 verde (l’urna α ) e 1 pallina rossa e 1 verde (l’urna β ).
Una prima osservazione, elementare ma molto utile, è che se un insieme A è in corrispondenza biunivoca con un insieme B, allora |A| = |B|. Un’altra osservazione, anch’essa molto intuitiva, è la seguente: se A, B sono due sottoinsiemi (di uno stesso spazio) disgiunti, cioè tali che A ∩ B = 0, / allora |A ∪ B| = |A| + |B|. Più in generale, se A1 , . . , Ak sono sottoinsiemi disgiunti, tali cioè che Ai ∩ A j = 0/ per i = j, allora | ki=1 Ai | = ∑ki=1 |Ai |. La dimostrazione di queste osservazioni è semplice ed è lasciata per esercizio.
Posto X = {1, 2, 3, 4, 5, 6}, prendiamo come spazio campionario Ω := X × X × X = {(x, y, z) : x, y, z ∈ N , 1 ≤ x, y, z ≤ 6} , munito della probabilità P uniforme. Si noti che A = {(x, y, z) ∈ Ω : x = y} , B = {(x, y, z) ∈ Ω : y = z} . Per il principio fondamentale del calcolo combinatorio, gli elementi di A sono determinati dalla scelta delle componenti x e z, per cui |A| = 6 · 6 = 36 e, analogamente, |B| = 36. Allo stesso modo, A ∩ B = {(x, y, z) ∈ Ω : x = y = z} e dunque |A ∩ B| = 6. Dato che |Ω | = |X|3 = 63 = 216, segue che P(A) = P(B) = 1 36 = , 216 6 P(A ∩ B) = |A ∩ B| 6 1 = = = P(A) P(B) , |Ω | 216 36 come richiesto.
Calcul des probabilites by Henri Poincaré
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