Final answer:
There are 330 ways to select disks with no bad sectors, 5 ways to select disks with all bad sectors, and 550 ways to select disks with exactly 2 having no bad sectors.
Step-by-step explanation:
The problem you've described is a combinatorial one, where we need to find the number of ways to select disks from a box under certain conditions, which relates to the combinations in probability theory.
A. Selection of Disks with No Bad Sectors
To select 4 disks that none have bad sectors from the 16 available, we need to choose from the 11 good disks. This can be calculated using the combination formula C(n, k) = n! / (k! * (n-k)!), where 'n' represents the number of items to choose from (good disks in this case), and 'k' is the number of items to choose (4 disks).
Combinations of choosing 4 good disks from 11 is C(11, 4) = 11! / (4! * (11-4)!) = 330 ways.
B. Selection of Disks with All Bad Sectors
When selecting 4 disks that all have bad sectors, we choose from the 5 bad disks. Using the combination formula, we get C(5, 4) = 5! / (4! * (5-4)!) = 5 ways.
C. Selection of Disks with Exactly 2 Good Disks
For selecting 4 disks where exactly 2 do not have bad sectors, we must choose 2 good disks from 11 and 2 bad disks from 5, and then multiply these combinations together.
Combinations for 2 good disks: C(11, 2) = 11! / (2! * (11-2)!) = 55 ways.
Combinations for 2 bad disks: C(5, 2) = 5! / (2! * (5-2)!) = 10 ways.
To get the total number of ways, we multiply these two results: 55 * 10 = 550 ways.