PHOTO TO NOTES: SOURCE-CHECKED PRACTICE PACK
ThetaWave editorial example | 2026-09-21
Article: https://thetawave.ai/blog/turn-photos-into-study-notes
This is fictional teaching material and an authored audit, not actual AI output,
a product accuracy test, or evidence of a workshop's effect on grades.

SOURCE S1 (readable transcript)
Uniformly select one student from the following fictional 200 students.
Group           Pass   Do not pass   Total
Workshop          60            20      80
No workshop       90            30     120
Total            150            50     200
W = attended the workshop. P = passed the test. | means given.
Find P(P | W), then P(W | P). Name the denominator group first.

A REUSABLE EXTRACTION INSTRUCTION (no guaranteed output)
Transcribe the title, every row and column heading, all cells and notation.
Keep the table structure. Mark unreadable text [unclear]; do not guess.
Then organize a study note. Label anything added beyond the source.
Keep source identifier S1 beside the table and calculations.

DELIBERATELY FLAWED DRAFT - DO NOT STUDY THIS VERSION
Workshop total = 60. P(P | W) = 60/200 = 30%.
P(W | P) = P(P | W) = 75%.

COMPLETED AUDIT
1. Row-label error: 60 is the workshop/pass cell, not its row total.
   S1 workshop row: 60 + 20 = 80. Correct the total to 80.
2. Condition omitted: 200 is everyone. Given W restricts the group to 80.
   Correct P(P | W) to 60/80 = 0.75 = 75%.
3. Condition reversed: given P restricts the group to the 150 who pass.
   Correct P(W | P) to 60/150 = 0.40 = 40%.

CHECKED STUDY NOTE
Source: S1. Goal: read the given group before choosing a denominator.
The table records counts, not percentages. Its row and column totals agree:
60+20=80; 90+30=120; 60+90=150; 20+30=50; 80+120=200.
Given W: use the workshop row. 60 of 80 pass, so P(P | W)=75%.
Given P: use the pass column. 60 of 150 attended, so P(W | P)=40%.
Added explanation: for a nonzero conditioning probability,
P(A | B)=P(A and B)/P(B). Among equally likely counted people, divide
by the number in group B. Reversing the condition can change the answer.
Added explanation reference: OpenStax Introductory Statistics 2e, section 3.3:
https://openstax.org/books/introductory-statistics-2e/pages/3-3-two-basic-rules-of-probability
This table alone cannot establish that a workshop causes test performance.

CLOSE THE NOTE: PRACTICE QUESTIONS
1. Using S1, given that the student did not attend, what is the probability
   that they pass? Name the denominator before calculating.
2. Using S1, given that the student passes, what is the probability that
   they did not attend the workshop?
3. New fictional table: workshop 48 pass/32 do not (80 total);
   no workshop 72 pass/48 do not (120 total). What is P(W | P)?

STOP HERE UNTIL YOU HAVE ATTEMPTED ALL THREE QUESTIONS.

ANSWER KEY
1. Denominator = 120 non-attendees. 90/120 = 75%.
2. Denominator = 150 students who pass. 90/150 = 60%.
3. Denominator = 48+72 = 120 students who pass. 48/120 = 40%.
   48/80 = 60% would answer the reversed condition P(P | W).

COMPLETED FICTIONAL REPAIR LOG
Attempt: question 3, answered 48/80 = 60%.
Error: selected workshop row although P (pass) is after the bar.
Repair: underline 'given pass'; use the pass column, 48+72=120.
Retry without key: 48/120 = 40%.
Scope: correct on this practice task; not proof of lasting mastery.

YOUR REUSABLE NOTE (copy this section)
Source name/image order:
Study goal:
Readable facts, table/diagram labels and units:
Unclear text to recapture (do not guess):
Checked note with source pointers:
Added explanation and separate reference:
Closed-source question:
My attempt:
Correct answer and reason:
Error to repair:
Next attempt: