Probabilistic Methods — Interactive Solver

University Homework & Passing

A survey of 100 students recorded whether each student $did\ homework$ (A) and whether they $passed$ the course (B). The joint counts are: A∩B = 48, A∩Bᶜ = 12, Aᶜ∩B = 22, Aᶜ∩Bᶜ = 18. Find P(A), P(B), P(A∪B), P(A|B), and P(B|A). Are A and B mutually exclusive? Are they independent?

Homework \ Result	Passed (B)	Failed (Bᶜ)	Total
Did homework (A)	48	12	60
No homework (Aᶜ)	22	18	40
Total	70	30	100

Step 1 — Define events and read the table

A = student did homework, B = student passed. Grand total n = 100. Row totals: P(A) row = 48+12 = 60; Aᶜ row = 22+18 = 40. Column totals: B col = 48+22 = 70; Bᶜ col = 12+18 = 30.

Step 2 — Marginal probabilities

Divide each marginal count by 100.

P(A) = \frac{60}{100} = 0.60, \quad P(B) = \frac{70}{100} = 0.70

Step 3 — Union via inclusion-exclusion

P(A∩B) = 48/100 = 0.48.

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.60 + 0.70 - 0.48 = 0.82

Step 4 — Conditional probability P(A|B)

Given the student passed, what fraction did homework?

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{48/100}{70/100} = \frac{48}{70} \approx 0.686

Step 5 — Conditional probability P(B|A)

Given the student did homework, what fraction passed?

P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{48}{60} = 0.80

Step 6 — Mutual exclusivity check

Mutually exclusive means A∩B = ∅, i.e. P(A∩B) = 0. Here P(A∩B) = 0.48 ≠ 0, so A and B are NOT mutually exclusive.

P(A \cap B) = 0.48 \neq 0

Step 7 — Independence check

Independent iff P(A|B) = P(A). We found P(A|B) ≈ 0.686 but P(A) = 0.60. They differ, so A and B are NOT independent — doing homework is associated with a higher pass rate.

P(A \mid B) \approx 0.686 \neq 0.60 = P(A)

Venn diagram

The four regions show joint counts out of 100: 48 students did homework AND passed ( $A \cap B$ ), 12 did homework but failed, 22 passed without doing homework, and 18 neither did homework nor passed.

Answer

P(A)=0.60,\ P(B)=0.70,\ P(A\cup B)=0.82,\ P(A\mid B)\approx 0.686,\ P(B\mid A)=0.80.\ \text{Not mutually exclusive; not independent.}

Explanation tip: Students who did homework passed at an 80% rate vs. the overall 70% rate, confirming that homework completion and passing are positively associated.

Support Tickets — First-Contact Resolution

350 support tickets are classified by type: Technical (T) or Non-technical (Tᶜ), and by whether they were resolved on first contact (S). Counts: T∩S = 90, T∩Sᶜ = 60, Tᶜ∩S = 160, Tᶜ∩Sᶜ = 40. Find P(T∪S), P(S|T), P(S|Tᶜ), and interpret what the difference says about technical tickets.

Ticket type \ Resolution	Resolved 1st contact (S)	Not resolved (Sᶜ)	Total
Technical (T)	90	60	150
Non-technical (Tᶜ)	160	40	200
Total	250	100	350

Step 1 — Read marginals from the table

Row totals: T = 90+60 = 150; Tᶜ = 160+40 = 200. Column totals: S = 90+160 = 250; Sᶜ = 60+40 = 100. Grand total = 350.

Step 2 — Marginal probabilities

P(T) = \frac{150}{350} \approx 0.429, \quad P(S) = \frac{250}{350} \approx 0.714

Step 3 — Union

P(T \cup S) = P(T) + P(S) - P(T \cap S) = \frac{150}{350} + \frac{250}{350} - \frac{90}{350} = \frac{310}{350} \approx 0.886

Step 4 — P(S|T): first-contact rate for technical tickets

P(S \mid T) = \frac{90}{150} = 0.60

Step 5 — P(S|Tᶜ): first-contact rate for non-technical tickets

P(S \mid T^c) = \frac{160}{200} = 0.80

Step 6 — Comparison and independence check

P(S|T) = 0.60 < P(S|Tᶜ) = 0.80, so being technical DECREASES the likelihood of first-contact resolution. T and S are not independent.

P(S \mid T) = 0.60 \neq 0.714 = P(S)

Venn diagram

Counts out of 350: 90 tickets are both Technical and resolved on first contact ( $T \cap S$ ), 60 are Technical but not resolved, 160 are Non-technical and resolved, and 40 are Non-technical and not resolved.

Answer

P(T\cup S)\approx 0.886,\ P(S\mid T)=0.60,\ P(S\mid T^c)=0.80.\ \text{Technical tickets are resolved on first contact less often.}

Explanation tip: Despite comprising only 43% of tickets, technical issues are harder to close immediately — a signal to invest in technical support training.

Online Course — Videos vs. Quiz Pass

Watched videos \ Quiz result	Passed (Q)	Failed (Qᶜ)	Total
Watched (W)	72	18	90
Did not watch (Wᶜ)	28	32	60
Total	100	50	150

Step 1 — Marginals

W total = 72+18 = 90; Wᶜ total = 28+32 = 60. Q total = 72+28 = 100; Qᶜ total = 18+32 = 50. Grand total = 150.

Step 2 — P(Q|W): pass rate among video watchers

P(Q \mid W) = \frac{72}{90} = 0.80

Step 3 — P(W|Q): fraction of passers who watched

P(W \mid Q) = \frac{72}{100} = 0.72

Step 4 — P(Q|Wᶜ): pass rate among non-watchers

P(Q \mid W^c) = \frac{28}{60} \approx 0.467

Step 5 — P(W|Qᶜ): fraction of failers who watched

P(W \mid Q^c) = \frac{18}{50} = 0.36

Step 6 — Interpretation of the two directions

P(Q|W) = 0.80 answers "does watching the videos help?" — 80% of watchers pass vs. ~47% of non-watchers. P(W|Q) = 0.72 describes the composition of passers — 72% of those who passed had watched. These are different conditional probabilities and must not be confused.

Venn diagram

Counts out of 150: 72 users both watched videos and passed ( $W \cap Q$ ), 18 watched but failed, 28 passed without watching, and 32 neither watched nor passed.

Answer

P(Q\mid W)=0.80,\ P(W\mid Q)=0.72,\ P(Q\mid W^c)\approx 0.467,\ P(W\mid Q^c)=0.36.

Explanation tip: Watching the videos nearly doubles the pass rate (80% vs. 47%). P(Q|W) measures the effect of watching; P(W|Q) measures who the successful students are — two very different questions.

Workplace Tool Adoption

200 employees are surveyed on tool usage: A = uses Tool A, B = uses Tool B. P(A) = 130/200, P(B) = 90/200, and 60 use both. Find the counts for each cell, then compute P(A∪B), the probability of using only A, only B, neither, P(A|B), and P(B|A).

Tool A \ Tool B	Uses B	Does not use B	Total
Uses A	60	70	130
Does not use A	30	40	70
Total	90	110	200

Step 1 — Derive cell counts

A total = 130, B total = 90, A∩B = 60. Only A = 130−60 = 70. Only B = 90−60 = 30. Neither = 200−130−30 = 40.

Step 2 — Union

P(A \cup B) = \frac{130 + 90 - 60}{200} = \frac{160}{200} = 0.80

Step 3 — Exclusive probabilities

P(A \text{ only}) = \frac{70}{200} = 0.35, \quad P(B \text{ only}) = \frac{30}{200} = 0.15, \quad P(\text{neither}) = \frac{40}{200} = 0.20

Step 4 — P(A|B)

P(A \mid B) = \frac{60}{90} = \frac{2}{3} \approx 0.667

Step 5 — P(B|A)

P(B \mid A) = \frac{60}{130} \approx 0.462

Step 6 — Independence check

Since P(A∩B) ≠ P(A)·P(B), the tools are not adopted independently.

P(A) \cdot P(B) = 0.65 \times 0.45 = 0.2925 \neq \frac{60}{200} = 0.30

Venn diagram

Counts out of 200: 70 employees use only Tool A, 60 use both ( $A \cap B$ ), 30 use only Tool B, and 40 use neither tool.

Answer

P(A\cup B)=0.80;\ \text{only A}=0.35,\ \text{only B}=0.15,\ \text{neither}=0.20;\ P(A\mid B)\approx 0.667,\ P(B\mid A)\approx 0.462.

Explanation tip: Two-thirds of Tool-B users also use Tool A, but fewer than half of Tool-A users also use Tool B — Tool A has broader reach in this workforce.

Streaming App — Independence Check

200 subscribers: A = has the mobile app installed, M = watches movies regularly. P(A) = 0.60, P(M) = 0.70, A∩M = 84. Build the full table and determine whether A and M are independent.

App installed \ Watches movies	Watches movies (M)	Does not watch (Mᶜ)	Total
Has app (A)	84	36	120
No app (Aᶜ)	56	24	80
Total	140	60	200

Step 1 — Derive cell counts

A total = 0.60×200 = 120; M total = 0.70×200 = 140; A∩M = 84. A∩Mᶜ = 120−84 = 36. Aᶜ∩M = 140−84 = 56. Aᶜ∩Mᶜ = 200−120−56 = 24.

Step 2 — P(M|A)

P(M \mid A) = \frac{84}{120} = 0.70

Step 3 — P(M|Aᶜ)

P(M \mid A^c) = \frac{56}{80} = 0.70

Step 4 — Independence criterion

P(M|A) = P(M|Aᶜ) = P(M) = 0.70. Also verify multiplicatively:

P(A) \cdot P(M) = 0.60 \times 0.70 = 0.42 = \frac{84}{200} = P(A \cap M) \checkmark

Step 5 — Conclusion

Both conditions confirm independence: knowing whether a subscriber has the app gives no information about whether they watch movies regularly.

Venn diagram

Counts out of 200: 84 subscribers both have the app and watch movies ( $A \cap M$ ), 36 have the app but do not watch, 56 watch movies without the app, and 24 have neither. Independence means $P(M \mid A) = P(M) = 0.70$ — the circle overlap is exactly proportional.

Answer

P(M\mid A)=P(M\mid A^c)=P(M)=0.70.\ A\ \text{and}\ M\ \text{are INDEPENDENT.}

Explanation tip: App installation and movie-watching habits are statistically independent — the platform is accessed equally across channels, suggesting no channel-specific engagement difference.

Delivery Delays — International vs. Domestic

600 parcels: I = international shipment, D = delayed. P(I) = 200/600, P(D) = 60/600, I∩D = 36. Build the table, compute P(D|I), P(D|Iᶜ), test independence, and find P(I|D).

Shipment type \ Delivery outcome	Delayed (D)	On time (Dᶜ)	Total
International (I)	36	164	200
Domestic (Iᶜ)	24	376	400
Total	60	540	600

Step 1 — Derive cell counts

I total = 200; D total = 60; I∩D = 36. I∩Dᶜ = 200−36 = 164. Iᶜ∩D = 60−36 = 24. Iᶜ∩Dᶜ = 400−24 = 376.

Step 2 — P(D|I): delay rate for international

P(D \mid I) = \frac{36}{200} = 0.18

Step 3 — P(D|Iᶜ): delay rate for domestic

P(D \mid I^c) = \frac{24}{400} = 0.06

Step 4 — Independence check

Not independent — international shipments are 3× more likely to be delayed.

P(D) = \frac{60}{600} = 0.10 \neq 0.18 = P(D \mid I)

Step 5 — Bayes: P(I|D)

Given a parcel is delayed, what is the probability it was international?

P(I \mid D) = \frac{P(I \cap D)}{P(D)} = \frac{36}{60} = 0.60

Venn diagram

Counts out of 600: 164 parcels are International but on time, 36 are International AND Delayed ( $I \cap D$ ), 24 are Domestic and Delayed, and 376 are Domestic and on time.

Answer

P(D\mid I)=0.18,\ P(D\mid I^c)=0.06,\ P(I\mid D)=0.60.\ \text{Not independent; international shipments are 3\times more delay-prone.}

Explanation tip: Although international shipments make up only 1/3 of volume, they account for 60% of all delays. Targeted logistics improvements for international routes would have the greatest impact.

Subscription Tiers — Law of Total Probability & Bayes

400 customers are split into tiers: High (100), Medium (150), Low (150). Renewal counts: High renews 80, Medium 90, Low 30. Use the law of total probability to find P(Renewed), then apply Bayes' theorem to find the tier probabilities given renewal.

Tier \ Renewal	Renewed	Not renewed	Total
High	80	20	100
Medium	90	60	150
Low	30	120	150
Total	200	200	400

Step 1 — Prior probabilities and conditional renewal rates

P(H)=\frac{100}{400}=0.25,\quad P(M)=\frac{150}{400}=0.375,\quad P(L)=\frac{150}{400}=0.375

Step 2 — Conditional renewal rates

P(R\mid H)=\frac{80}{100}=0.80,\quad P(R\mid M)=\frac{90}{150}=0.60,\quad P(R\mid L)=\frac{30}{150}=0.20

Step 3 — Law of total probability

P(R) = P(H)\cdot P(R\mid H)+P(M)\cdot P(R\mid M)+P(L)\cdot P(R\mid L)

Step 3 (calculation)

P(R) = 0.25\times 0.80+0.375\times 0.60+0.375\times 0.20 = 0.20+0.225+0.075 = 0.50

Step 4 — Bayes: P(H|R)

P(H\mid R)=\frac{P(H)\cdot P(R\mid H)}{P(R)}=\frac{0.25\times 0.80}{0.50}=\frac{0.20}{0.50}=0.40

Step 5 — Bayes: P(M|R)

P(M\mid R)=\frac{0.375\times 0.60}{0.50}=\frac{0.225}{0.50}=0.45

Step 6 — Bayes: P(L|R)

P(L\mid R)=\frac{0.375\times 0.20}{0.50}=\frac{0.075}{0.50}=0.15

Step 7 — Verify

0.40+0.45+0.15=1.00 \checkmark

Probability tree

Each leaf shows the joint probability $P(\text{Tier}) \times P(R \mid \text{Tier})$ . The three Renew leaves sum to $P(R) = 0.20 + 0.225 + 0.075 = 0.50$ .

Law of total probability

Answer

P(R)=0.50;\ P(H\mid R)=0.40,\ P(M\mid R)=0.45,\ P(L\mid R)=0.15.

Explanation tip: Even though Medium and Low tiers have the same count, renewers skew heavily toward High and Medium. Low-tier customers renew at only 20%, making them the primary churn risk despite representing 37.5% of the base.

Fraud Detection — Base-Rate Paradox

10 000 transactions: 100 are fraudulent (F), 9 900 are legitimate (Fᶜ). A detection system flags suspicious transactions (S) with P(S|F) = 0.98 and P(S|Fᶜ) = 0.03. Build the table, compute P(S), and find P(F|S) — the probability a flagged transaction is actually fraud.

Transaction \ Flagged?	Flagged (S)	Not flagged (Sᶜ)	Total
Fraud (F)	98	2	100
Legitimate (Fᶜ)	297	9603	9900
Total	395	9605	10000

Step 1 — Derive expected counts

Fraud: 100 × 0.98 = 98 flagged, 2 not flagged. Legitimate: 9900 × 0.03 = 297 flagged, 9603 not flagged.

Step 2 — P(S) via law of total probability

P(S)=P(F)\cdot P(S\mid F)+P(F^c)\cdot P(S\mid F^c)

Step 2 (calculation)

P(S)=0.01\times 0.98+0.99\times 0.03=0.0098+0.0297=0.0395

Step 3 — Bayes: P(F|S)

P(F\mid S)=\frac{P(F)\cdot P(S\mid F)}{P(S)}=\frac{0.0098}{0.0395}\approx 0.248

Step 4 — Interpret the result

Even with a 98% true-positive rate, only about 24.8% of flagged transactions are genuinely fraudulent. The remaining ~75% are false positives from the large pool of legitimate transactions.

Step 5 — Verify from table

\frac{98}{98+297}=\frac{98}{395}\approx 0.248 \checkmark

Probability tree

$P(F \mid S) = 0.0098 / (0.0098 + 0.0297) \approx 0.248$ — most flagged transactions are false positives from the much larger pool of legitimate transactions.

Bayes tree (fraud detection)

Venn diagram

Per 10 000 transactions: 2 are Fraud but not flagged, 98 are Fraud AND flagged ( $F \cap S$ ), 297 are Legit but flagged, and 9 603 are Legit and not flagged.

Answer

P(S)=0.0395;\ P(F\mid S)\approx 0.248.\ \text{Only ~24.8\% of flagged transactions are truly fraudulent.}

Explanation tip: This is the classic base-rate (false-positive) paradox: a rare event (1% fraud rate) means even a highly sensitive detector produces mostly false alarms. Posterior probability depends critically on the prior prevalence.

Order Cancellations by Channel

An e-commerce platform processes orders via three channels: Web (50% of orders, 4% cancel), App (35%, 6% cancel), Phone (15%, 10% cancel). Per 1 000 orders, build the table and find the overall cancellation probability P(C). Then use Bayes to find which channel is most responsible for cancellations.

Channel \ Order outcome	Cancelled	Completed	Total
Web	20	480	500
App	21	329	350
Phone	15	135	150
Total	56	944	1000

Step 1 — Per-1000 counts

Web: 500 orders → 500×0.04 = 20 cancelled, 480 completed. App: 350 orders → 350×0.06 = 21 cancelled, 329 completed. Phone: 150 orders → 150×0.10 = 15 cancelled, 135 completed.

Step 2 — Law of total probability

P(C)=P(\text{Web})\cdot P(C\mid \text{Web})+P(\text{App})\cdot P(C\mid \text{App})+P(\text{Phone})\cdot P(C\mid \text{Phone})

Step 2 (calculation)

P(C)=0.50\times 0.04+0.35\times 0.06+0.15\times 0.10=0.020+0.021+0.015=0.056

Step 3 — P(Web|C)

P(\text{Web}\mid C)=\frac{0.020}{0.056}=\frac{20}{56}\approx 0.357

Step 4 — P(App|C)

P(\text{App}\mid C)=\frac{0.021}{0.056}=\frac{21}{56}=0.375

Step 5 — P(Phone|C)

P(\text{Phone}\mid C)=\frac{0.015}{0.056}=\frac{15}{56}\approx 0.268

Step 6 — Verify and interpret

The App channel contributes the most cancellations (37.5%) despite Phone having the highest individual cancellation rate (10%), because App handles 35% of volume versus Phone's 15%.

0.357+0.375+0.268=1.000 \checkmark

Probability tree

Each leaf is the joint probability $P(\text{channel}) \times P(C \mid \text{channel})$ . The three Cancel leaves sum to $P(C) = 0.020 + 0.021 + 0.015 = 0.056$ .

Cancellation by channel

Answer

P(C)=0.056;\ P(\text{Web}\mid C)\approx 0.357,\ P(\text{App}\mid C)=0.375,\ P(\text{Phone}\mid C)\approx 0.268.

Explanation tip: The App channel is responsible for the most cancellations in absolute terms. Even though phone orders cancel most often per order, the app's higher volume makes it the dominant source — a volume-weighted view changes the priority for intervention.

Onboarding Tutorial — Comprehensive Analysis

500 users: T = completed the onboarding tutorial, C = completed the product setup. Counts: T∩C = 180, T∩Cᶜ = 70, Tᶜ∩C = 120, Tᶜ∩Cᶜ = 130. Compute all marginals, the union, four conditional probabilities, check independence, and quantify the effect of the tutorial.

Tutorial \ Setup completion	Completed (C)	Not completed (Cᶜ)	Total
Did tutorial (T)	180	70	250
Skipped tutorial (Tᶜ)	120	130	250
Total	300	200	500

Step 1 — Marginals

T total = 180+70 = 250; Tᶜ total = 120+130 = 250. C total = 180+120 = 300; Cᶜ total = 70+130 = 200. Grand total = 500.

P(T)=\frac{250}{500}=0.50,\quad P(C)=\frac{300}{500}=0.60

Step 2 — Union

P(T\cup C)=P(T)+P(C)-P(T\cap C)=0.50+0.60-\frac{180}{500}=0.50+0.60-0.36=0.74

Step 3 — P(C|T): completion rate with tutorial

P(C\mid T)=\frac{180}{250}=0.72

Step 4 — P(C|Tᶜ): completion rate without tutorial

P(C\mid T^c)=\frac{120}{250}=0.48

Step 5 — P(T|C): fraction of completers who did tutorial

P(T\mid C)=\frac{180}{300}=0.60

Step 6 — P(T|Cᶜ): fraction of non-completers who did tutorial

P(T\mid C^c)=\frac{70}{200}=0.35

Step 7 — Independence check

T and C are NOT independent. Tutorial completion raises the setup completion rate from 48% to 72% — a 24 percentage-point lift.

P(C\mid T)=0.72 \neq 0.60=P(C)

Step 8 — Effect size summary

The tutorial is associated with a 24 pp increase in setup completion probability.

P(C\mid T)-P(C\mid T^c)=0.72-0.48=0.24

Venn diagram

Counts out of 500: 70 users did the tutorial but did not complete setup, 180 did both ( $T \cap C$ ), 120 completed setup without the tutorial, and 130 did neither.

Answer

P(T)=0.50,\ P(C)=0.60,\ P(T\cup C)=0.74;\ P(C\mid T)=0.72,\ P(C\mid T^c)=0.48,\ P(T\mid C)=0.60,\ P(T\mid C^c)=0.35.\ \text{Not independent; tutorial lifts completion by +24 pp.}

Explanation tip: Users who did the tutorial complete setup at 72% vs. 48% for those who skipped it. This 24 percentage-point lift is strong evidence that the onboarding tutorial meaningfully improves activation — a compelling case for making it more prominent.

List 6 — Conditional Probability

University Homework & Passing

Support Tickets — First-Contact Resolution

Online Course — Videos vs. Quiz Pass

Workplace Tool Adoption

Streaming App — Independence Check

Delivery Delays — International vs. Domestic

Subscription Tiers — Law of Total Probability & Bayes

Fraud Detection — Base-Rate Paradox

Order Cancellations by Channel

Onboarding Tutorial — Comprehensive Analysis