Question

A box contains 16 computer disks, 5 of which are known to have bad sectors. In how many ways can 4 disks be selected, without replacement and without regard to order, so that the following conditions are satisfied? A. In how many ways can disks be selected so that none have bad sectors? B. In how many ways can disks be selected so that all have bad sectors? C. In how many ways can disks be selected so that exactly 2 do not have bad sectors?

Answer 1

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.