Probability of survival of one disk, p=1-0.05=0.95
If there are n drives, and failures are independent (e.g. mechanical wear and tear, and not a lightning strike or a current spike), then
Probability of survival = p^n
(a) for two disks, P(2) = 0.95^2=0.9025
(b)for three disks P(3) = 0.95^3 = 0.8574
Division is the inverse of the multiplication.
Answer:
can you do elaborate please
Let's see
![\sf [0.5\:0.5]\left[\begin{array}{cc}\sf b_1 &\sf r_1\\ \sf b_2&\sf r_2\end{array}\right]](https://tex.z-dn.net/?f=%5Csf%20%5B0.5%5C%3A0.5%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Csf%20b_1%20%26%5Csf%20r_1%5C%5C%20%5Csf%20b_2%26%5Csf%20r_2%5Cend%7Barray%7D%5Cright%5D)
r_1 and r2 are costs per pound
So
- It's r_1 is the cost of 1/2 pound rice at market one and r_2 is at market 2
Option J