OpenCyber Academy
Rapid course · Lesson 13 · English
Course code: OCA-012Lesson 13: OCA-012-A013

Three of twelve counts conjunctions

Open, free learning for anyone who wants to protect people and technology. No account is required to read this text or its worked answers.

Lesson map

Lesson 13. Map, example, EX059.

flowchart TD
  C["46 ones inside 60 cells"] --> R["A cell rate"]
  U["Nine laptops with at least one zero"] --> W["A row rate"]
  R --> N{"Are 46/60 and 3/12 the same claim?"}
  W --> N
  N -->|No| K["Report the reduction you computed"]

Cell totals and row totals answer different questions. Their failure sets overlap.

Logical drawing

pass all five          FB01, FB08, FB12
update zeros           FB03, FB07
encryption zeros       FB02, FB06, FB07, FB11
standard-user zeros    FB02, FB04, FB07, FB09
lock zeros             FB05
backup zeros           FB03, FB07, FB10
union of those zeros   nine keys, not fourteen

Learning outcomes

After this chapter, you should be able to:

  1. Compute the cell total, the conjunction, and the union from the dictionary.
  2. Explain why fourteen failure-cells are not fourteen laptops.
  3. Refuse a kappa, an average disguised as the all-five count, or a tie-break this run did not need.

13.1 Overlapping failure sets are why the rates do not add.

The five printed numerators are 10, 8, 8, 11, and 9. Their sum is 46. Each laptop contributes five cells, and twelve laptops contribute 60 cells. So 46 of 60 cells are 1. That ratio is a property of cells. It is not 3/12. The all-five count asks a harder question of each row: are all five cells 1? Only FB01, FB08, and FB12 meet it. A row with a single 0 contributes four, or even five, to a careless reading of "mostly fine," and contributes nothing to 3/12. Both arithmetic results are in the object. Substituting one for the other changes the question.

The failure side shows why adding the column gaps over-counts laptops. Update fails on {FB03, FB07}. Encryption fails on {FB02, FB06, FB07, FB11}. The standard-user cell fails on {FB02, FB04, FB07, FB09}. The lock fails on {FB05}. Backup fails on {FB03, FB07, FB10}. Those sets contain 2, 4, 4, 1, and 3 keys, and 2+4+4+1+3 is 14. The union contains nine keys: FB02, FB03, FB04, FB05, FB06, FB07, FB09, FB10, and FB11. Nine plus the three clean rows is twelve. The inequality is ordinary. A sum of sizes equals the size of the union only when the sets are disjoint. They are not. FB07 sits in four of the five failure sets and only passes the lock, so its five-cell sum is 1. FB02 fails encryption and the standard-user cell. FB03 fails updates and backup. Adding 14 and announcing "14 laptops failed" invents two laptops and hides the overlap.

min selects FB07 because 1 is the unique minimum. Python dictionaries preserve insertion order as a language guarantee from 3.7 onward, yet a tie never arose, so that rule stayed idle. Landis and Koch define kappa for agreement between observers on categorical labels. One person filled this matrix. A second observer was not used, and a kappa was not computed. Do not decorate 46 of 60 with a coefficient the study did not run. The later narrative in which twelve of twelve pass is still a different reading. It is not a reinterpretation of these nine keys.

13.2 Worked example

Lesson table; scroll horizontally to read all columns →

Quantity Value from this dictionary Decision
Yes-cells 46 of 60 A cell total
All-five rows 3 of 12 A conjunction
Sum of column-failure sizes 14 Not a laptop count
Union of failing keys 9 The rows that miss the conjunction

Synthetic case. The sets are the zeros already visible in the embedded lists.

Errors and limits

An average of the five column rates is still not the conjunction. Overlap is the reason the story feels inconsistent until the sets are written down.

Apply

  • EX059 — Which total? A summary says "about three quarters pass, so 9 of 12 pass all five," and it adds a kappa. What does the dictionary actually support?
    Show marking criteria
    Rubric: 3 = 46 of 60 is the cell total; 3 of 12 is the conjunction on FB01, FB08 and FB12; column-failure sizes sum to 14 across a union of nine keys; FB07's sum is 1 and there is no tie; no kappa is computed;; 2 = the cell total and the row total stay distinct;; 1 = one total;; 0 = none.
Worked answers

EX059. 46 of 60 is the cell total. 3 of 12 is the conjunction on FB01, FB08 and FB12. Failure sizes sum to 14 across nine keys. FB07's sum is 1, with no tie. No kappa is computed.