Answer:
A) 0 , B) 0.0199 , C) 200
Explanation:
Let X denotes no. of computers fail on day
X ~ Bin (n = 4 , p = 0.05)
A] P (X = x) = (n c x) p^x q^n-x
P (X = 4) = (4 c 4) (0.005)^4 (0.995)^0
(0.05)^4 ~ 0
B] P (X > 1) = 1 - P (X < 1) = 1 - P (X = 0)
1 - { ( 4 c 0) (0.005)^0 (0.995)^ 4 }
1 - (0.995)^ 4 ~ 0.0199
C] Let Y depict mean number of days until a specific computer fails
Y ~ Geom (p = 0.005)
Mean = E (Y) = 1/p = 1/0.05 = 200