Conditional Probability Flashcards: Two-Way Table Practice
Practice conditional probability with 44 original cards: choose the given group, calculate table ratios, reverse conditions, and repair denominator mistakes.
Bu deste hakkında
Practice conditional probability with 44 original English flashcards built around small invented count tables. Each card asks for one result: a given group, an exact probability fraction, a needed count, a corrected setup, or a short interpretation. Row-given and column-given questions, a few reversed conditions, conditional complements, and joint and marginal contrasts make denominator choices visible. Exact group rates support short count reconstruction and one reverse-condition calculation. Limit cards address empty groups, zero overlaps, missing information, unequal selection weights, rounded rates, and causal claims.
The opening four cards establish notation, overlap, the count-ratio rule, and complete table categories. The remaining sequence mixes calculations, interpretations, count reconstruction, and error repair, introducing more demanding rates and limits later. Repeated-table variants are separated by at least three cards on other tasks; the review scheduler handles long-term spacing. Every prompt includes its own needed givens. Count-to-probability exercises state uniform item selection.
This is a compact recall and error-repair companion for high-school and introductory statistics learners. It excludes a broad Bayes theorem formula curriculum, continuous distributions, inference tests, independence proof drills, medical diagnosis, full course or exam coverage, full worksheet construction, and every forward/reverse permutation. These would require other practice formats or a wider scope. Recalling these fixed examples does not establish mastery of unfamiliar probability problems. Learn the method with conditional probability two-way table practice, then solve fresh problems with different counts. The statistics flashcard study guide gives broader review-planning context.
The prompts, invented counts, explanations, sequence, metadata, and original cover were independently created with AI assistance and reviewed for factual and arithmetic accuracy. Mathematical rules were checked against Penn State’s conditional probability lessons and the University of Minnesota’s two-way-table reference; no source exercises, wording, tables, figures, or exam material were copied. Original elements are released under CC0 1.0 to the extent applicable rights exist; third-party expression is excluded. This general topic deck claims no official examination alignment, affiliation, or endorsement.
Bu destedeki kartlar
Kart 1
Soru
In P(A | B), which event is the given condition?
Cevap
B. Restrict attention to outcomes in B; A is the event whose probability is requested.
Kart 2
Soru
When one item is selected uniformly from a finite collection, what count ratio gives P(A | B)?
Cevap
Count in both A and B / count in B, provided the B count is positive. In probability notation, P(A | B) = P(A and B) / P(B), with P(B) > 0.
Kart 3
Soru
What does the event “A and B” include?
Cevap
Only outcomes that satisfy both A and B. It is their overlap, also written A ∩ B; it need not include every outcome in B.
Kart 4
Soru
For a complete two-way count table, how many cells should each item contribute to?
Cevap
One cell. Each variable’s categories must be mutually exclusive and exhaustive, so no item is duplicated or omitted.
Kart 5
Soru
Choose one token uniformly from this complete table. Each item appears once.
Dot No dot Blue 6 9 Not blue 4 11What is P(dot | blue)?
Cevap
2/5. The blue group has 6 + 9 = 15 tokens; 6 have dots. Use 6/15 = 2/5.
Kart 6
Soru
Choose one ticket uniformly from this complete table. Each item appears once.
Stamp No stamp Striped 9 6 Plain 3 12What is P(striped | stamped)?
Cevap
3/4. There are 9 + 3 = 12 stamped tickets; 9 are striped. Use 9/12 = 3/4.
Kart 7
Soru
Choose one paper flower uniformly from this complete table. Each item appears once.
Folded Unfolded Red 5 7 Not red 10 8A learner writes P(folded | red) = 5/15. What denominator should replace 15?
Cevap
12. The condition is red, so use the red row total, 5 + 7. The 15 counts folded flowers instead.
Kart 8
Soru
A complete batch has 40 silver and 20 non-silver badges. Exactly 3/5 of the silver badges are numbered; all other silver badges are unnumbered. How many badges are both silver and numbered?
Cevap
24 badges. Apply the within-silver rate to the silver total: (3/5) × 40 = 24.
Kart 9
Soru
Choose one book uniformly from this complete table. Each item appears once.
Borrowed On shelf Hardback 8 10 Paperback 15 7What is P(borrowed | hardback)?
Cevap
4/9. The hardback group contains 8 + 10 = 18 books; 8 are borrowed. Use 8/18 = 4/9.
Kart 10
Soru
In a complete registration list, P(completed | online) = 3/4 under uniform selection. What does 3/4 describe in context?
Cevap
Three quarters of the online registrations are completed. The fraction describes the online group, not all registrations or the group of completed registrations.
Kart 11
Soru
Choose one tile uniformly from this complete table. Each item appears once.
Marked Unmarked Round 7 3 Not round 5 9What is P(round | marked)?
Cevap
7/12. The marked column has 7 + 5 = 12 tiles; 7 are round.
Kart 12
Soru
Choose one token uniformly from this complete table. Each item appears once.
Dot No dot Blue 6 9 Not blue 4 11What is P(blue | dot)?
Cevap
3/5. The dotted group has 6 + 4 = 10 tokens; 6 are blue. Use 6/10 = 3/5.
Kart 13
Soru
Choose one folder uniformly from this complete table. Each item appears once.
Tagged Untagged New 4 6 Old 8 12A learner calls 4/10 the probability of “new and tagged.” What fraction answers that joint-probability question?
Cevap
2/15. “New and tagged” means 4 folders out of all 30: 4/30 = 2/15. The 4/10 ratio conditions on new.
Kart 14
Soru
Choose one bead uniformly from this complete table. Each item appears once.
Holed Solid Green 0 8 Not green 5 7What is P(holed | green)?
Cevap
0. The green group is nonempty: 0 + 8 = 8. No green bead is holed, so the ratio is 0/8 = 0.
Kart 15
Soru
Choose one cup uniformly from this complete table. Each item appears once.
Handle No handle Blue 9 6 Not blue 3 12What is P(no handle | blue)?
Cevap
2/5. Of the 15 blue cups, 6 have no handle. Use 6/15 = 2/5; the condition stays blue.
Kart 16
Soru
In a complete collection of envelopes, 18 are both small and sealed. Exactly 3/4 of all small envelopes are sealed. How many small envelopes are there?
Cevap
24 envelopes. If n is the small-group total, (3/4) × n = 18, so n = 18 ÷ (3/4) = 24.
Kart 17
Soru
Choose one button uniformly from this complete table. Each item appears once.
Hole No hole Green 6 4 Not green 4 12What is P(hole | green)?
Cevap
3/5. The green row has 6 + 4 = 10 buttons; 6 have holes. Use 6/10 = 3/5.
Kart 18
Soru
A complete note collection gives P(blue | sealed) = 2/7 under uniform selection. Which group is the denominator of this fraction?
Cevap
All sealed notes. The numerator counts notes that are both blue and sealed.
Kart 19
Soru
A complete collection has 40 large and 30 small labels. Exactly 1/4 of the large labels are stamped; all other large labels are unstamped. How many large labels are unstamped?
Cevap
30 labels. There are (1/4) × 40 = 10 stamped large labels; 40 − 10 = 30 are unstamped.
Kart 20
Soru
Choose one book uniformly from this complete table. Each item appears once.
Borrowed On shelf Hardback 8 10 Paperback 15 7What is P(hardback | borrowed)?
Cevap
8/23. The borrowed group has 8 + 15 = 23 books; 8 are hardbacks. The 18-book hardback row is not the given group.
Kart 21
Soru
A batch has 60 small and 40 large boxes, with no other sizes. Exactly 1/5 of small boxes and 3/10 of large boxes are marked; the rest are unmarked. Choose one box uniformly. What is P(small | marked)?
Cevap
1/2. There are (1/5) × 60 = 12 marked small boxes and (3/10) × 40 = 12 marked large boxes. The marked group has 24 boxes, so use 12/24.
Kart 22
Soru
Choose one label uniformly from this complete table. Each item appears once.
Stamped Unstamped Purple 0 0 Not purple 6 10What is P(stamped | purple) under the count-ratio rule?
Cevap
Undefined. The purple group is empty, so the ratio would be 0/0. This rule requires a positive conditioning-group count.
Kart 23
Soru
Choose one tile uniformly from this complete table. Each item appears once.
Marked Unmarked Round 7 3 Not round 5 9What is P(not round | unmarked)?
Cevap
3/4. The unmarked group has 3 + 9 = 12 tiles; 9 are not round. Use 9/12 = 3/4.
Kart 24
Soru
Choose one parcel uniformly from this complete table. Each item appears once.
Sealed Open Small 6 9 Large 10 5A learner writes P(small) = 6/16. What fraction gives P(small), with no condition?
Cevap
1/2. All 15 small parcels count, out of 30 total: 15/30 = 1/2. The 6/16 ratio is P(small | sealed).
Kart 25
Soru
Choose one order uniformly from this complete table. Each item appears once.
Printed Unprinted Early 11 4 Not early 7 8What is P(printed | not early)?
Cevap
7/15. The not-early row has 7 + 8 = 15 orders; 7 are printed.
Kart 26
Soru
A complete collection has 80 cards. Exactly 3/8 of all cards are both blue and dotted. How many cards are both blue and dotted?
Cevap
30 cards. The rate uses the whole collection: (3/8) × 80 = 30.
Kart 27
Soru
For uniform selection from a complete pin collection, P(round | gold) = 1/4. A learner labels 3/4 as P(round | not gold). Which conditional probability does 3/4 actually give?
Cevap
P(not round | gold). Take the complement of the requested event while keeping the condition gold: 1 − 1/4 = 3/4. The non-gold group can have a different rate.
Kart 28
Soru
In a nonempty red-tag group, exactly 2/3 are square. No other counts or rates are supplied. Can you determine the fraction of square tags that are red?
Cevap
No. You need the count that is both red and square and the total square count, or equivalent proportions. The within-red rate alone does not determine the reverse condition.
Kart 29
Soru
Choose one notebook uniformly from this complete table. Each item appears once.
Lined Blank Small 2 6 Medium 5 3 Large 3 1What is P(medium | lined)?
Cevap
1/2. The lined column contains 2 + 5 + 3 = 10 notebooks; 5 are medium. Use 5/10.
Kart 30
Soru
Uniform selection from a complete list of 50 workshop attendees gives P(brings pen | morning session) = 4/5. To which population does this result directly apply?
Cevap
The morning-session attendees in that 50-person list. This fraction alone does not establish the rate for attendees at other workshops.
Kart 31
Soru
Choose one tile uniformly from this complete table. Each item appears once.
Dot No dot Blue 5 7 Not blue 4 8What is P(dot), without a given color?
Cevap
3/8. There are 5 + 4 = 9 dotted tiles among 24 total. Use 9/24 = 3/8.
Kart 32
Soru
For uniform selection from a collection of tags, P(square | red) = 2/5. A learner says, “Two fifths of the square tags are red.” What group should replace “square tags”?
Cevap
The red tags. The result says two fifths of red tags are square. It does not give P(red | square).
Kart 33
Soru
Choose one ribbon uniformly from this complete table. Each item appears once.
Dot No dot Red 4 6 Blue 2 3 Yellow 6 9A learner uses 4/(4 + 2) for P(red | dot). What fraction corrects the missing category?
Cevap
1/3. Include every dotted ribbon: 4 + 2 + 6 = 12. Use 4/12 = 1/3.
Kart 34
Soru
A complete tray has 10 red and 10 non-red pieces. The selector is more likely to choose each red piece than each non-red piece. Can counts alone justify P(red) = 10/20?
Cevap
No. The count ratio needs uniform item selection. With unequal selection probabilities, use the pieces’ selection weights; equal counts do not imply equal chances.
Kart 35
Soru
Choose one basket uniformly from this complete table. Each item appears once.
Lid No lid Round 4 10 Not round 9 5What is P(no lid | round)?
Cevap
5/7. The round row contains 4 + 10 = 14 baskets; 10 have no lid. Use 10/14 = 5/7.
Kart 36
Soru
Under uniform selection from a complete collection of items, P(sealed | blue) = 3/5 and P(blue | sealed) = 1/2. Which group does the 1/2 rate describe?
Cevap
The sealed items. Half of that group is blue; the 3/5 rate instead describes the blue group.
Kart 37
Soru
A complete list contains only morning (20) and afternoon (50) entries. Exactly 10 in each group are checked; the rest are unchecked. A learner says the checked fractions are equal because both counts are 10. Which group has the larger checked fraction?
Cevap
The morning group. Its fraction is 10/20 = 1/2, while the afternoon fraction is 10/50 = 1/5. Compare within-group rates, not raw counts.
Kart 38
Soru
A complete collection has 25 red and 15 non-red tokens. Exactly 18 tokens have dots, including 8 red tokens. Every other token has no dot. How many non-red tokens have dots?
Cevap
10 tokens. The dotted column splits into red and non-red: 18 − 8 = 10.
Kart 39
Soru
Choose one entry uniformly from this complete table. Each item appears once.
Checked Unchecked Bus 6 4 Walk 3 7 Cycle 9 1What is P(not bus | checked)?
Cevap
2/3. The checked group contains 6 + 3 + 9 = 18 entries. The walk and cycle entries give 3 + 9 = 12 non-bus entries, so use 12/18 = 2/3.
Kart 40
Soru
In an observational club list, 4/5 of members who attend a workshop finish a project, versus 2/5 of those who do not attend. Does this comparison alone prove that workshop attendance causes project completion?
Cevap
No. The conditional rates describe an association in that list. Other differences between the groups can explain it; the comparison alone does not establish causation.
Kart 41
Soru
Choose one bookmark uniformly from this complete table. Each item appears once.
Folded Flat Orange 7 5 Not orange 3 9What is P(orange and folded)?
Cevap
7/24. The joint event contains 7 bookmarks out of all 24. Neither a row nor a column is specified as a given condition.
Kart 42
Soru
Choose one card uniformly from this complete table. Each item appears once.
Star No star Blue 8 6 Not blue 5 11What count belongs in the denominator of P(blue | star)?
Cevap
13. The given group is the star column: 8 + 5 = 13 cards.
Kart 43
Soru
A complete group has 300 items. Its marked percentage is reported as 33%, rounded to the nearest whole percent. Must exactly 99 items be marked?
Cevap
No. A rounded rate does not fix an exact count. Both 98/300 ≈ 32.67% and 99/300 = 33% round to 33%.
Kart 44
Soru
Choose one chip uniformly from this complete table. Each item appears once.
Marked Unmarked Red 3 5 Not red 4 8A learner writes P(marked | red) = 8/3. What fraction corrects the inverted ratio?
Cevap
3/8. There are 3 marked red chips out of 3 + 5 = 8 red chips. A conditional probability cannot exceed 1.
44 kart
Conditional Probability Flashcards: Two-Way Table Practice
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