Calcul des probabilites by Henri Poincaré

By Henri Poincaré

Excerpt from Calcul des Probabilités

Le premiere 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 l. a. 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 l. a. 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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Chapter 3 39 Theoretical Exercises 1. P(ABA) = 2. If A ⊂ B P( AB ) P( AB ) ≥ = P(ABA ∪ B) P( A) P( A ∪ B) P( A) , P(ABc) = 0, P( B) P(AB) = 3. P(BA) = 1, P(BAc) = P( BAc ) P( Ac ) Let F be the event that a first born is chosen. Also, let Si be the event that the family chosen in method a is of size i. Pa(F) = 1 ni ∑ P( F S ) P( S ) = ∑ i m i i i Pb(F) = i m i ini ∑ Thus, we must show that ∑ in ∑ n / i ≥ m i i i 2 i or, equivalently, ∑ in ∑ n i i j /j≥ j ∑n ∑n i i j j or, i ∑∑ j n n ≥ ∑∑ n n i i≠ j j i j i≠ j Considering the coefficients of the term ninj, shows that it is sufficient to establish that i j + ≥2 j i or equivalently i2 + j2 ≥ 2ij which follows since (i − j)2 ≥ 0.

Assume n > 1. 2 (a) n (b) Conditioning on whether the man of couple j sits next to the woman of couple i gives the 1 1 n−2 2 2n − 3 result: + = n − 1 n − 1 n − 1 n − 1 (n − 1) 2 (c) e−2 68. exp(−10e−5} 69. 2212 54 Chapter 4 70. e−λt + (1 − e−λt)p 71.  26  (a)    38  5 3  26  12 (b)    38  38 72. 4) 3   73. Let N be the number of games played. 8125 74. 2 (a)   3 5 5 3 6 2 7 8  8  2   1   8  2   1   8  2   1   2  (b)      +      +      +    5  3   3   6  3   3   7  3   3   3  5  5  2  1 (c)     4  3  3  6  2  (d)     4  3  5 1   3 2 76.

E[Y] = E[X/σ − µ/σ] = 1 σ E[ X ] − µ/σ = µ/σ − µ/σ = 0 Var(Y) = (1/σ)2 Var(X) = σ2/σ2 = 1. n 10. E[1/(X + 1)] = i =0 n = ∑ i =0 = = 1 n! i! p (1 − p) i n −i n! (i + 1)! n  n + 1 1 (n + 1) p ∑  i + 1  p 1 ( n + 1) p ∑  i +1 (1 − p) n − i i =0 n +1  n + 1 j n +1− j  p (1 − p ) j  j =1   n + 1 0 1 n +1− 0  1 −  0  p (1 − p)  (n + 1) p     1 = [1 − (1 − p ) n +1 ] (n + 1) p = 11. For any given arrangement of k successes and n − k failures: P{arrangementtotal of k successes} = 12.

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