Bayes’ Theorem Flashcards: Base Rates & Evidence Updates
Practice 48 original Bayes theorem flashcards: build source-and-evidence weights, normalize across all sources, repair base-rate mistakes, and check sequential updates.
Despre acest pachet
A signal can be common for one source without making that source likely after the signal. These 48 original Bayes theorem flashcards help you keep the source share, source-specific likelihood, total evidence, and posterior separate.
Start with priors and likelihoods. Then calculate prior × likelihood branch weights for fictional bags, batches, routes, and sensors; add every possible source branch to get the evidence total; and normalize the requested source weight. Two-source updates lead into three-source comparisons and subset posteriors. Short repair prompts target reversed conditions, missing priors or branches, the wrong complement, and branch weights mistaken for posteriors. Six cards handle sequential evidence, including when conditional independence permits multiplication and when a repeated display supplies no new observation.
The sequence introduces each prerequisite before its dependent calculation, then interleaves calculations, interpretations, and repairs. Related variants are spaced so one answer does not directly cue the next; the app’s scheduler handles later review spacing. Each card gives the model assumptions it needs and an explained answer.
This is a focused introductory probability set, not a full statistics or exam course. It avoids two-way-table count drills already covered by the separate conditional-probability deck, mechanical reverse calculations that cannot recover unknown inputs, continuous Bayesian models, odds-form algebra, medical and financial decisions, and claims about real-world sources from toy assumptions. All examples, wording, order, and cover are original. Mathematical facts were checked against NIST’s Engineering Statistics Handbook and MIT OCW’s probability lecture; their prose, exercises, and figures were not copied. AI assisted drafting, followed by individual factual and numerical review. The CC0 label covers only the original expression and generated artwork to the extent rights exist.
Fișele din acest pachet
Fișa 1
Întrebare
For a Bayes update over hidden sources, what must the listed sources cover?
Răspuns
Every possible source, exactly once. The sources must be exhaustive and mutually exclusive so their evidence branches can be added.
Fișa 2
Întrebare
Before seeing a result, a token comes from bag A with probability 0.3. What is 0.3 called in a Bayes update?
Răspuns
The prior for bag A. It is A’s source probability before the result is observed.
Fișa 3
Întrebare
A token from bag A is blue with probability 0.8. In a Bayes update after seeing blue, what role does 0.8 play?
Răspuns
The likelihood of blue given A. It is P(blue | A), not the chance that a blue token came from A.
Fișa 4
Întrebare
An observed signal can come from every possible source. What probability belongs in the denominator of a source posterior?
Răspuns
The total probability of the signal. Add its prior × likelihood branch weight across all possible sources.
Fișa 5
Întrebare
Only bags A and B can be chosen. P(A) = 0.3 and P(red | A) = 0.8. What is the joint weight for A and red?
Răspuns
0.24. Multiply 0.3 × 0.8; this is P(A and red), before normalization.
Fișa 6
Întrebare
A bag yields blue with probability 0.8. A learner calls 0.8 the chance the bag was chosen after seeing blue. What is the mistake?
Răspuns
The condition was reversed. 0.8 is P(blue | bag); P(bag | blue) also needs source priors and the other bags’ blue rates.
Fișa 7
Întrebare
Only batches A and B supply parts. Their shares are 0.25 and 0.75; a part is marked with probabilities 0.6 and 0.2, respectively. What is P(marked)?
Răspuns
0.30. Add both marked branches: 0.25 × 0.6 + 0.75 × 0.2 = 0.15 + 0.15.
Fișa 8
Întrebare
A source makes 70% of its tokens striped and supplies 20% of all tokens. A learner uses 0.70 as its striped branch weight. What factor is missing?
Răspuns
The 0.20 prior. Its striped branch weight is 0.20 × 0.70 = 0.14.
Fișa 9
Întrebare
Only routes A and B are possible. Their priors are 0.4 and 0.6; P(chime | A) = 0.9 and P(chime | B) = 0.1. After a chime, what is P(A | chime)?
Răspuns
6/7. A contributes 0.36; both routes contribute 0.36 + 0.06 = 0.42. Divide 0.36/0.42.
Fișa 10
Întrebare
Only A and B exist. A learner divides A’s bell branch weight by itself and calls the result P(A | bell). What denominator term is missing?
Răspuns
P(B)P(bell | B). The bell denominator must include B’s branch too, if B can produce a bell.
Fișa 11
Întrebare
Only bins A and B exist. A’s share rises from 0.1 to 0.4 while P(glow | A) = 0.8 and P(glow | B) = 0.2 stay fixed. Does P(A | glow) rise or fall?
Răspuns
It rises. The posterior changes from 4/13 to 8/11 because the prior for A increased with both likelihoods fixed.
Fișa 12
Întrebare
Only sources A, B, and C exist. Their priors are 0.2, 0.3, 0.5; their beep rates are 0.5, 0.4, 0.2. What is P(beep)?
Răspuns
0.32. Add every branch: 0.10 + 0.12 + 0.10.
Fișa 13
Întrebare
For a source A, a learner uses P(A) × P(A) × P(glow | A) as its glow branch weight. What is the correction?
Răspuns
Use the prior once. The branch weight is P(A) × P(glow | A).
Fișa 14
Întrebare
Only sources A and B exist. A supplies 35% of items. What prior belongs on B’s branch?
Răspuns
65%. The two source shares must add to 100%.
Fișa 15
Întrebare
After a red result, the only three source-and-red weights are A: 0.12, B: 0.08, C: 0.20. What is P(A | red)?
Răspuns
0.3. The total red weight is 0.40, so normalize A’s 0.12/0.40.
Fișa 16
Întrebare
Only A and B are equally likely. Given each source, two new signals are independent. Their likelihoods are 0.8 and 0.6 under A, and 0.2 and 0.3 under B. After both signals, what is P(A | both)?
Răspuns
8/9. Joint weights are 0.5 × 0.8 × 0.6 = 0.24 for A and 0.5 × 0.2 × 0.3 = 0.03 for B.
Fișa 17
Întrebare
Only routes North and South exist. North is used 70% of the time, and P(late | North) = 10%. What is the North-and-late branch weight?
Răspuns
7%. Multiply 70% × 10%; it is not the posterior P(North | late).
Fișa 18
Întrebare
Sources A, B, C have priors 0.1, 0.3, 0.6 and signal likelihoods 0.9, 0.5, 0.4. Which source has the largest signal branch weight?
Răspuns
C. The weights are 0.09, 0.15, and 0.24; A has the largest likelihood but not the largest contribution.
Fișa 19
Întrebare
A model gives signal rates for sources A and B but says a third source may exist, without its share or signal rate. Can P(A | signal) be determined?
Răspuns
No. The evidence total needs every possible source branch; an unmodelled source leaves the denominator incomplete.
Fișa 20
Întrebare
Only sources Z and Y are possible. Their priors are 0.3 and 0.7; P(ring | Z) = 0.4 and P(ring | Y) = 0.1. After a ring, what is P(Z | ring)?
Răspuns
12/19. The branch weights are 0.12 and 0.07; normalize Z’s weight by 0.19.
Fișa 21
Întrebare
Only machines A and B produce tokens. Their shares are 0.6 and 0.4, and their stripe rates are 0.25 and 0.5. What is P(stripe)?
Răspuns
0.35. Add 0.6 × 0.25 and 0.4 × 0.5, giving 0.15 + 0.20.
Fișa 22
Întrebare
After a pulse, the only source-and-pulse weights are A: 0.10, B: 0.20, C: 0.30. What is P(A or B | pulse)?
Răspuns
0.5. A and B contribute 0.30 of the total 0.60 pulse weight.
Fișa 23
Întrebare
After one observation, P(A | first result) = 0.7. A genuinely new second result arrives. What prior for A enters the next conditional update?
Răspuns
0.7. The first posterior becomes the starting probability for the new evidence; its likelihood still needs a model.
Fișa 24
Întrebare
After a click, the only branch weights are A: 0.10 and B: 0.30. A learner says P(A | click) = 0.75. What is the corrected value?
Răspuns
0.25. Normalize A’s weight, 0.10/(0.10 + 0.30); 0.75 belongs to B.
Fișa 25
Întrebare
Only bags A and B are possible. Their priors are 0.2 and 0.8; P(star | A) = 0.75 and P(star | B) = 0.25. After seeing a star, what is P(A | star)?
Răspuns
3/7. A’s star weight is 0.15 and total star weight is 0.15 + 0.20 = 0.35.
Fișa 26
Întrebare
Only sources A, B, C have priors 0.25, 0.25, 0.5 and tick likelihoods 0.4, 0.2, 0.3. A proposed P(tick) is 0.15 because it omits C. What is the corrected P(tick)?
Răspuns
0.30. Add C’s 0.5 × 0.3 = 0.15 to the A and B weights, 0.10 + 0.05.
Fișa 27
Întrebare
Only A and B exist and their priors are known. P(ping | A) is known, but P(ping | B) is missing. Is P(A | ping) uniquely determined?
Răspuns
No. B’s unknown ping contribution changes the denominator.
Fișa 28
Întrebare
Only A and B make tiles. P(A) = 0.2, P(B) = 0.8, and P(star | A) = 0.7. A learner uses 0.8 × 0.7 for A’s star branch. What is A’s correct branch weight?
Răspuns
0.14. A’s prior is 0.2, so multiply 0.2 × 0.7. The 0.8 prior belongs to B.
Fișa 29
Întrebare
A learner wants P(not A | ring) but subtracts P(A | no ring) from 1. Which probability should be subtracted from 1?
Răspuns
P(A | ring). Keep the observed condition fixed when taking the source complement.
Fișa 30
Întrebare
After a stripe, the only three source-and-stripe weights are A: 0.06, B: 0.09, C: 0.15. What is P(C | stripe)?
Răspuns
0.5. Divide C’s 0.15 by the total 0.30.
Fișa 31
Întrebare
A sensor result appears twice because the screen refreshed; no second measurement occurred. Should a Bayes update use the same result twice?
Răspuns
No. A duplicated display supplies no new evidence. Updating twice would count one observation twice.
Fișa 32
Întrebare
Only A and B exist. P(A) = 0.1, P(B) = 0.9, P(chime | A) = 0.9, and P(chime | B) = 0.4. A learner chooses A as more likely after a chime because 0.9 > 0.4. Which source is actually more likely?
Răspuns
B. The chime weights are A: 0.09 and B: 0.36; priors change the comparison.
Fișa 33
Întrebare
Only boxes A and B exist. P(A) = 0.3, and both boxes yield a square token with probability 0.4. After seeing a square, what is P(A | square)?
Răspuns
0.3. Equal likelihoods leave the prior unchanged: 0.3 × 0.4 / 0.4 = 0.3.
Fișa 34
Întrebare
Three mutually exclusive sources cover every possibility. After observing a flash with positive probability, what must their three posterior probabilities sum to?
Răspuns
1. Normalization divides every source-and-flash weight by the same positive total.
Fișa 35
Întrebare
Two signals are independent overall, but their dependence given each hidden source is unknown. Can their source-specific likelihoods safely be multiplied?
Răspuns
No. Multiplication for a source needs independence conditional on that source, or a supplied joint likelihood.
Fișa 36
Întrebare
After a pulse, the only two source-and-pulse weights are 0.40 and 0.10. A learner reports the first source’s posterior as 0.40. What is it?
Răspuns
0.8. Divide 0.40 by the total pulse weight 0.50; a branch weight is not yet a posterior.
Fișa 37
Întrebare
Only trays A and B exist and are equally likely. Their dot rates are 0.6 and 0.3. After seeing a dot, what is P(A | dot)?
Răspuns
2/3. The weights are 0.30 and 0.15; A contributes 0.30/0.45.
Fișa 38
Întrebare
Only sensors A, B, C are possible. Their priors are 0.5, 0.3, 0.2 and ping likelihoods are 0.1, 0.4, 0.5. After a ping, what is P(C | ping)?
Răspuns
10/27. C contributes 0.10; all sources contribute 0.05 + 0.12 + 0.10 = 0.27.
Fișa 39
Întrebare
A model assigns an observed flash probability of zero across every possible source. Is P(A | flash) defined by the ordinary Bayes ratio?
Răspuns
No. Its denominator P(flash) is zero, so the conditional probability is undefined in this discrete model.
Fișa 40
Întrebare
Only sources A and B exist. A supplies 5% of items; P(flash | A) = 0.9 and P(flash | B) = 0.1. What is P(flash)?
Răspuns
0.14. Add 0.05 × 0.9 and 0.95 × 0.1, giving 0.045 + 0.095.
Fișa 41
Întrebare
Only A, B, C have priors 0.1, 0.4, 0.5 and mark likelihoods 0.9, 0.1, 0.1. Which source is most probable after a mark?
Răspuns
A. Its weight 0.09 exceeds B’s 0.04 and C’s 0.05, despite A’s smallest prior.
Fișa 42
Întrebare
Signals are independent given source A, but nothing is stated about their dependence given source B. Can both sources’ two-signal branch weights be built by multiplying separate likelihoods?
Răspuns
No. The B branch still needs P(both signals | B) or conditional independence given B.
Fișa 43
Întrebare
A proposed evidence total adds source A, source B, and a third label that includes every B item. What is the structural error?
Răspuns
B is counted twice. The branches used in a total must be disjoint and cover all possible sources.
Fișa 44
Întrebare
An observed click has branch weights 0.12 from source A and 0.18 from source B; A and B are the only sources. What is P(A | click)?
Răspuns
0.4. Normalize A’s weight: 0.12/(0.12 + 0.18).
Fișa 45
Întrebare
After a glow, the only source-and-glow weights are A: 0.08, B: 0.12, C: 0.20. What is P(not C | glow)?
Răspuns
0.5. A and B contribute 0.20 of the 0.40 total; the complement is taken after conditioning on glow.
Fișa 46
Întrebare
A classroom model assigns fixed bag shares and color rates. Does a posterior computed from it establish the source of a real token outside that model?
Răspuns
No. The result is conditional on the stipulated shares, rates, and source list; a real setting needs checked assumptions and data.
Fișa 47
Întrebare
Only sources A and B exist with priors 0.4 and 0.6. Supplied joint likelihoods for two signals are P(both | A) = 0.12 and P(both | B) = 0.24. What is P(A | both)?
Răspuns
0.25. The weights are 0.4 × 0.12 = 0.048 and 0.6 × 0.24 = 0.144; divide 0.048 by 0.192.
Fișa 48
Întrebare
Only sources A and B exist. P(A) = 0.4, P(light | A) = 0.8, and P(light | B) = 0.3. After no light, what is P(A | no light)?
Răspuns
4/25. No-light weights are 0.4 × 0.2 = 0.08 and 0.6 × 0.7 = 0.42; divide 0.08 by 0.50.
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Bayes’ Theorem Flashcards: Base Rates & Evidence Updates
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