Exercice 2:

Question 1: MTBF = 392 donc lambda = 1 / 392 = 0.00255
Question 2: F(365) = P(T <= 365) = 1 - e(-0.00255*365) = 0.61
Question 3:
	3 années = 3 * 365 = 1095
	R(1095) = P(T > 1095) = e(-0.00255*1095) = 0.06
Question 4:
		P(T > n) >= 0.95
	<=> e(-0.00255*n) >= 0.95
	<=> -0.00255*n >= ln(0.95)
	<=> n <= ln(0.95)/(-0.00255)
	<=> n <= 20.11 soit n <= 20
Question 5:
	a) F(365)^6 = (1 - e(-0.00255*365))^6 = 0.05
	b) 1 - F(365)^6 = 1 - 0.05 = 0.95