da vl4d » 12/09/2006, 19:05
la 1 potrebbe essere $(5!)^2/(10!) ≈ 0.0039$ ?
per la 2, diciamo che $X_{i} = 1$ se la coppia $i$ e' italiana, 0 altrimenti. Abbiamo al massimo 5 coppie italiane dunque $i$ va da 0 a 4, abbiamo $P(X_{i}) = (5-i)/(10-i)$ dunque
$E[X] = \sum_{i=0}^4 X_{i}P_{X_{i}} = 1.78$
Ma aspettiamo cheguevilla...
Go to the roots, of these calculations! Group the operations.
Classify them according to their complexities rather than their appearances!
This, I believe, is the mission of future mathematicians. This is the road on which I am embarking in this work.
Evariste Galois