NCERT Plus One (Class 11) Mathematics • Chapter 14

Probability

Q: Find the probability that a leap year selected at random will contain 53 Sundays.

Key (1 Mark): 366 days = 52 weeks (52 Sundays) + 2 extra consecutive days . Possible pairs: (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon) (7 pairs). 2 pairs include Sunday ⇒ Probability = 2 7 .

Q: Check whether P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.6 are consistently defined.

Key (1 Mark): NO (Inconsistent). Since A ∩ B ⊆ A , we must have P(A ∩ B) ≤ P(A) , whereas here 0.6 > 0.5 (also giving P(A − B) = −0.1 < 0 , impossible!).

Q: A coin is tossed. If it shows head, we draw a ball from a bag containing 3 blue and 4 white balls; if it shows tail, we throw a die. Describe the sample space.

Let blue balls be B₁, B₂, B₃ and white balls W₁, W₂, W₃, W₄ . S = {HB₁, HB₂, HB₃, HW₁, HW₂, HW₃, HW₄, T1, T2, T3, T4, T5, T6} . Total outcomes: n(S) = 7 + 6 = 13 .

Q: One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card will be (i) a diamond, (ii) not an ace.

n(S) = 52 . • (i) 13 diamonds ⇒ P(diamond) = 13 52 = 1 4 . • (ii) 4 aces ⇒ P(ace) = 4 52 = 1 13 ⇒ P(not ace) = 1 − 1 13 = 12 13 .

Q: Given P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B), (ii) P(A' ∩ B'), (iii) P(A ∩ B').

• (i) P(A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88 . • (ii) P(A' ∩ B') = P((A ∪ B)') = 1 − 0.88 = 0.12 . • (iii) P(A ∩ B') = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19 .

Q: In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both. If one student is selected at random, find the probability that the student opted for: (i) NCC or NSS, (ii) neither NCC nor NSS, (iii) NSS but not NCC.

P(NCC) = 30/60, P(NSS) = 32/60, P(NCC ∩ NSS) = 24/60 . • (i) P(NCC ∪ NSS) = 30 + 32 − 24 60 = 38 60 = 19 30 . • (ii) P(neither) = 1 − 19 30 = 11 30 . • (iii) P(NSS − NCC) = 32 − 24 60 = 8 60 = 2 15 .

Q: A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man, (b) one man, (c) two men?

Total ways = ⁴C₂ = 6 . • (a) No man (2 women): ²C₂ ⁶ = 1 6 . • (b) One man & one woman: ²C₁ × ²C₁ 6 = 4 6 = 2 3 . • (c) Two men: ²C₂ 6 = 1 6 . (Check: 1/6 + 2/3 + 1/6 = 1 !)

Q: Two students Anil and Ashima appeared in an exam. P(Anil qualifies) = 0.05, P(Ashima qualifies) = 0.10, and P(both qualify) = 0.02. Find P that: (a) both will not qualify, (b) at least one will not qualify, (c) only one of them will qualify.

Let E = Anil qualifies, F = Ashima qualifies. P(E ∪ F) = 0.05 + 0.10 − 0.02 = 0.13 . • (a) Both not: P(E' ∩ F') = 1 − P(E ∪ F) = 1 − 0.13 = 0.87 . • (b) At least one not: P(E' ∪ F') = 1 − P(E ∩ F) = 1 − 0.02 = 0.98 . • (c) Only one: P(E) + P(F) − 2P(E ∩ F) = 0.05 + 0.10 − 2(0.02) = 0.11 .

Random experiments, sample spaces, event algebra, mutually exclusive vs exhaustive events, Kolmogorov's axioms & addition/complement laws.

📄 Official NCERT Textbook PDF

Topic 14.1 – 14.1.2: Random Experiments, Outcomes, Sample Space & Compound Events

Right before a school cricket match in Kochi, the captain flips a coin. Everyone knows the two possible results — Head (H) or Tail (T) — yet nobody can predict the exact result ahead of time! Any experiment whose full menu of outcomes is known, while individual results stay unpredictable, is a Random Experiment. That complete menu is the Sample Space (S), and every subset you care about is an Event (E ⊆ S)!

SAMPLE SPACE (S)Set of ALL possible outcomes (n coins → 2ⁿ; n dice → 6ⁿ)

SIMPLE vs COMPOUNDSimple: 1 outcome {HH} | Compound: 2+ outcomes {HT, TH}

SURE vs IMPOSSIBLES = Sure event (always occurs) | ∅ = Impossible event (never occurs!)

🧠 Memory Trick: Think of Sample Space S as the entire ice-cream menu card, and an Event E as your chosen subset order! 1 scoop = Simple; 2+ scoops = Compound!

💡 Simple Kid-Check: Roll one die: the Sample Space is S = {1, 2, 3, 4, 5, 6}. Asking for "an even number" picks E = {2, 4, 6} (3 outcomes → Compound Event). Asking for "number 7" picks ∅ (Impossible Event)!

1
Tossing Coins: 1 coin → {H, T} (2); 2 coins → {HH, HT, TH, TT} (4); 3 coins → 2³ = 8 outcomes.
2
Rolling Two Dice: Ordered pairs (x, y) with x, y ∈ {1..6} give 6 × 6 = 36 sample points!
3
Branching Experiments: Toss a coin; if Head roll a die, if Tail toss coin again → S = {H1, H2, H3, H4, H5, H6, TH, TT} (8 outcomes!).

Event Type	Subset Size in S	Example (Rolling One Die)
Impossible Event	`n(E) = 0` (`∅`)	Getting a multiple of 7: `∅`
Simple (Elementary)	`n(E) = 1`	Getting exact number 5: `{5}`
Compound Event	`n(E) ≥ 2`	Getting a prime number: `{2, 3, 5}`
Sure Event	`E = S`	Getting a number < 7: `{1,2,3,4,5,6}`

Watch TWO classic counting traps! (1) Writing S = {HH, HT, TT} (only 3 outcomes!) for two coins — NO! First coin Head + second Tail (HT) is completely distinct from TH, so 2 coins ALWAYS have 4 outcomes! (2) Saying 2 dice have 6 + 6 = 12 outcomes — outcomes MULTIPLY (6 × 6 = 36), never add!

Topic 14.1.3 – 14.1.5: Mutually Exclusive vs Exhaustive Events (Set Algebra Mapping)

Roll one die: can a single roll land BOTH Odd (A = {1, 3, 5}) AND Even (B = {2, 4, 6}) at the exact same instant? IMPOSSIBLE! Their overlap is empty (A ∩ B = ∅), so they are Mutually Exclusive. Plus, combining them fills all 6 faces (A ∪ B = S), so they are Exhaustive too! Probability events speak pure Set Algebra!

EVENT ALGEBRAA or B = A ∪ B | A and B = A ∩ B | A not B = A − B

MUTUALLY EXCLUSIVECannot happen together: A ∩ B = ∅ (zero overlap!)

EXHAUSTIVE EVENTSCover ALL of S together: E₁ ∪ E₂ ∪ ... ∪ Eₙ = S

🧠 Memory Trick: Exclusive = Empty Intersection (∩ = ∅); Exhaustive = Full Union (∪ = S)! Remember: two events can be Exclusive WITHOUT being Exhaustive (e.g. {1,2} and {5,6} on a die)!

💡 Simple Kid-Check: Chapter 1 Sets returns as your superpower! Whenever a question says OR, reach for Union (∪). Whenever it says AND, reach for Intersection (∩). Whenever it says NOT, reach for Complement (')!

English Event Statement	Set Notation	Die Example (`A = {2,3,5}, B = {1,3,5}`)
Event "A or B"	`A ∪ B`	`{1, 2, 3, 5}`
Event "A and B"	`A ∩ B`	`{3, 5}` (not exclusive!)
Event "A but not B"	`A − B = A ∩ B'`	`{2}`
Event "not A" / "neither A nor B"	`A'` / `A' ∩ B' = (A ∪ B)'`	`A' = {1, 4, 6}` / `(A∪B)' = {4, 6}`

🎲 NCERT Check (Rolling One Die, S = {1,2,3,4,5,6}):
• A = {1, 2, 3}, B = {3, 4}, C = {5, 6}: Since A ∪ B ∪ C = S, they are Exhaustive! Yet A ∩ B = {3} ≠ ∅, so A and B are NOT Mutually Exclusive!
• E₁ = {1, 2} and E₂ = {5, 6}: E₁ ∩ E₂ = ∅ (Mutually Exclusive!), yet E₁ ∪ E₂ ≠ S (NOT Exhaustive!).

Top conceptual blunder in Section 14.1: assuming "Mutually Exclusive" automatically guarantees "Exhaustive" — FALSE! Check BOTH tests independently: (1) Pairwise intersections Eᵢ ∩ Eⱼ = ∅ for Exclusive, AND (2) Full union ⋃ Eᵢ = S for Exhaustive! Also note: all simple events of S are automatically mutually exclusive and exhaustive!

Topic 14.2: Axiomatic Approach to Probability (0 ≤ P(E) ≤ 1, Sum = 1)

What if a cheeky friend brings a weighted coin to a Kozhikode carrom match where Heads shows 70% of the time (0.7) and Tails shows 30% (0.3)? Counting equal outcomes fails for unfair coins! In 1933, Russian mathematician A. N. Kolmogorov built three crystal-clear Axioms assigning probability weights to sets — working smoothly for EVERY fair OR biased experiment!

3 KOLMOGOROV AXIOMS(1) P(E) ≥ 0 | (2) P(S) = 1 | (3) If E∩F=∅, P(E∪F) = P(E)+P(F)

OUTCOME WEIGHTSFor S = {ω₁, ..., ωₙ}: 0 ≤ P(ωᵢ) ≤ 1 AND Σ P(ωᵢ) = 1

EVENT PROBABILITYP(A) = Σ P(ωᵢ) for ωᵢ ∈ A | P(∅) = 0

🧠 Memory Trick: Imagine P(S) = 1 as one whole ₹1 coin shared among all outcomes: NO outcome gets negative money (< 0), NO outcome gets over ₹1, and ALL pieces MUST total exactly 1.00!

💡 Simple Kid-Check: Kolmogorov's Axioms act like a referee checking probability tables! Any negative number? Red card! Any number above 1? Red card! Total of all outcomes in S equal to 0.9 or 1.2 instead of 1? Red card!

1
Rule 1 (Non-negative): Every P(ωᵢ) ∈ [0, 1] (never negative, never > 1).
2
Rule 2 (Total = 1): Sum of probabilities of ALL elementary outcomes in S MUST equal 1.
3
Rule 3 (Disjoint Additivity): For disjoint events, probabilities add; taking F = ∅ proves P(∅) = 0!

🔍 NCERT Validity Check on S = {ω₁, ω₂, ω₃, ω₄}:
• (0.1, 0.2, 0.3, 0.4) → All in [0, 1] and sum = 1.0 → VALID!
• (−0.1, 0.5, 0.4, 0.2) → Sums to 1.0, YET −0.1 < 0 → INVALID!
• (1/6, 1/6, 1/6, 1/6) → Sum = 4/6 ≠ 1 → INVALID!

Watch TWO sneaky 1–2 mark DHSE traps! (1) Checking ONLY whether weights add to 1 and missing a negative value like −0.1! BOTH conditions (0 ≤ P(ωᵢ) ≤ 1 AND Σ = 1) must pass! (2) Consistency check: Can P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6? NO! Because A ∩ B ⊆ A forces P(A ∩ B) ≤ P(A), yet 0.6 > 0.5!

Topic 14.2.2: Equally Likely Outcomes & Playing Cards / Dice / Coin Models

When a coin, die, or well-shuffled deck of 52 playing cards is completely fair, every outcome shares equal probability 1n(S). Suddenly probability turns into pure counting: P(E) = n(E) / n(S)! Whether picking Red Kings from a 52-card pack or forming a 2-person committee using ⁿCᵣ, one simple fraction rules them all!

FAIR FORMULAP(E) = n(E)n(S) (ONLY when outcomes are equally likely!)

52-CARD DECK26 Black (13 ♠ Spade, 13 ♣ Club) + 26 Red (13 ♥ Heart, 13 ♦ Diamond)

FACE CARDS & ACES12 Face Cards (4 J, 4 Q, 4 K) | 4 Aces (Aces are NOT face cards!)

🧠 Memory Trick: Inside EVERY suit of 13 cards: 1 Ace + 9 Number cards (2–10) + 3 Face cards (J, Q, K) = 13! And for Two Dice Sums 2..12, counts form a perfect pyramid: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1!

💡 Simple Kid-Check: Why does P(E) = n(E)/n(S) work for a fair deck? Because all 52 cards have equal chance 1/52, so adding 1/52 for each of the n(E) winning cards gives n(E)/52! For an unfair die, NEVER use n(E)/n(S)!

52-Card Category	Count	Probability	Two-Dice Sum	Ways out of 36
Each Suit (♠, ♣, ♥, ♦)	`13`	`13/52 = 1/4`	Sum `2` or `12`	`1` each
Red or Black	`26`	`26/52 = 1/2`	Sum `6` or `8`	`5` each
Face Cards (J, Q, K)	`12`	`12/52 = 3/13`	Sum `7` (Peak!)	`6` (`6/36 = 1/6`)
Red King / Black Ace	`2`	`2/52 = 1/26`	Doublets `(x=y)`	`6` (`6/36 = 1/6`)

🃏 Classic NCERT Card & Combination Models:
• Card drawn is Red OR King: 26 Red + 4 Kings − 2 Red Kings = 28 ⇒ 2852 = 713!
• Committee of 2 from 2 Men & 2 Women (NCERT Ex 8): Total = ⁴C₂ = 6. P(1 man, 1 woman) = ²C₁ × ²C₁⁴C₂ = 46 = 23!

Watch THREE common exam traps! (1) Counting Aces as Face Cards — ONLY Jack, Queen, and King are Face Cards (12 total, NOT 16!). (2) Using n(E)/n(S) on a loaded die (e.g. a die with two faces marked 1, three faces 2, one face 3: P(2) = 3/6 = 1/2, NOT 1/3!). (3) Double-counting Red Kings in "Red or King"!

Topic 14.2.3: Addition Theorem: P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

In a Kollam Plus One batch, 50% of students play Football (A), 40% play Cricket (B), and 15% play BOTH (A ∩ B). What is the chance a random student plays Football OR Cricket? Adding 0.50 + 0.40 = 0.90 double-counts the dual-sport kids sitting in the middle overlap! Subtract that overlap once: 0.50 + 0.40 − 0.15 = 0.75!

ADDITION THEOREMP(A ∪ B) = P(A) + P(B) − P(A ∩ B)

MUTUALLY EXCLUSIVEIf A ∩ B = ∅, P(A ∩ B) = 0 ⇒ P(A ∪ B) = P(A) + P(B)

ONLY A / EXACTLY ONEP(A − B) = P(A) − P(A ∩ B) | P(only one) = P(A) + P(B) − 2P(A ∩ B)

🧠 Memory Trick: Why subtract 2 P(A ∩ B) for "Only ONE of A or B"? Because P(A) + P(B) counts the middle lens TWICE, while "only one" wants the middle lens ZERO times!

💡 Simple Kid-Check: Picture a Venn diagram made of three separate non-overlapping pieces: Only A (A − B), Both (A ∩ B), and Only B (B − A)! Adding P(A) + P(B) accidentally stacks the middle piece twice, so subtracting P(A ∩ B) once restores the exact Union!

4 Golden Addition & De Morgan Identities:
1) P(A or B) = P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
2) P(A but not B) = P(A ∩ B') = P(A) − P(A ∩ B)
3) P(neither A nor B) = P(A' ∩ B') = 1 − P(A ∪ B)
4) 3-Event Rule: P(A∪B∪C) = P(A)+P(B)+P(C) − P(A∩B) − P(B∩C) − P(C∩A) + P(A∩B∩C)

🎓 NCERT Example 7 (Anil & Ashima): Given P(E) = 0.05, P(F) = 0.10, P(E ∩ F) = 0.02:
• P(E ∪ F) = 0.05 + 0.10 − 0.02 = 0.13 (at least one qualifies)
• Both will NOT qualify: P(E' ∩ F') = 1 − P(E ∪ F) = 1 − 0.13 = 0.87
• Only ONE qualifies: (0.05 − 0.02) + (0.10 − 0.02) = 0.03 + 0.08 = 0.11!

Watch TWO frequent 3–4 mark exam traps! (1) Using P(A) + P(B) blindly when events are NOT mutually exclusive! (2) Mixing up "at least one of A or B" (P(A ∪ B) = P(A)+P(B)−P(A∩B)) with "only one of A or B" (P(A)+P(B)−2P(A∩B))! Notice the factor of 2!

Topic 14.2.4 & Misc: Complement Rule P(not A) = 1 − P(A) & 'At-Least-One' Tricks

Suppose you toss a fair coin 3 times and want the probability of getting "AT LEAST ONE Head". Counting 1H, 2H, 3H forces you to list 7 outcomes! Instead, ask: what is the ONLY outcome with ZERO Heads? Just TTT (1/8)! Subtract from 1: P(at least one H) = 1 − 1/8 = 7/8! Flipping tough questions upside down saves minutes in exams!

COMPLEMENT RULEP(A') = P(not A) = 1 − P(A) (since A ∪ A' = S & A ∩ A' = ∅)

AT LEAST ONEP(at least one) = 1 − P(NONE) (1-step shortcut!)

AT LEAST ONE NOTP(A' ∪ B') = 1 − P(A ∩ B) (De Morgan's flip!)

🧠 Memory Trick: Spot the magic trigger words "at least one" or "not both" → immediately flip to 1 − P(opposite single event)!

💡 Simple Kid-Check: Because an event A and its opposite A' split the entire universe into two non-overlapping halves, P(A) + P(A') = 1 ALWAYS! Whenever counting A needs 5 cases and counting A' needs just 1 case, ALWAYS compute 1 − P(A')!

1
Spot the Opposite ("None"): Opposite of "at least 1 prize in 10 tickets" is "0 prizes in all 10 tickets".
2
Compute Easy Case: For 10,000 lottery tickets with 10 prizes, P(no prize with 2 tickets) = ⁹⁹⁹⁰C₂¹⁰⁰⁰⁰C₂.
3
Flip via 1 − P(...): Also remember De Morgan's twin: P(neither) = 1 − P(A ∪ B) vs P(at least one fails) = 1 − P(A ∩ B)!

✉️ Classic NCERT Misc Q6 (3 Letters in 3 Addressed Envelopes):
Total random arrangements = 3! = 6 (ABC, ACB, BAC, BCA, CAB, CBA).
How many put ZERO letters in the right envelope? Only 2 (BCA, CAB)!
Therefore P(at least 1 in right envelope) = 1 − 26 = 46 = 23!

Watch the word-for-word exam trap between these two phrases (NCERT Example 7!):
• "BOTH A and B do NOT happen" = P(A' ∩ B') = 1 − P(A ∪ B)
• "AT LEAST ONE of A or B does NOT happen" = P(A' ∪ B') = 1 − P(A ∩ B)!
Reading carelessly swaps ∪ and ∩ and costs 2 marks!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the probability that a leap year selected at random will contain 53 Sundays.

Key (1 Mark): 366 days = 52 weeks (52 Sundays) + 2 extra consecutive days.
Possible pairs: (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon) (7 pairs).
2 pairs include Sunday ⇒ Probability = 27.

Q2 (Kerala HSE Say • 1 Marks): Check whether P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.6 are consistently defined.

Key (1 Mark): NO (Inconsistent).
Since A ∩ B ⊆ A, we must have P(A ∩ B) ≤ P(A), whereas here 0.6 > 0.5 (also giving P(A − B) = −0.1 < 0, impossible!).

Q3 (Kerala HSE March • 2 Marks): A coin is tossed. If it shows head, we draw a ball from a bag containing 3 blue and 4 white balls; if it shows tail, we throw a die. Describe the sample space.

Let blue balls be B₁, B₂, B₃ and white balls W₁, W₂, W₃, W₄.
S = {HB₁, HB₂, HB₃, HW₁, HW₂, HW₃, HW₄, T1, T2, T3, T4, T5, T6}.
Total outcomes: n(S) = 7 + 6 = 13.

Q4 (Kerala HSE Model • 2 Marks): One card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card will be (i) a diamond, (ii) not an ace.

n(S) = 52.
• (i) 13 diamonds ⇒ P(diamond) = 1352 = 14.
• (ii) 4 aces ⇒ P(ace) = 452 = 113 ⇒ P(not ace) = 1 − 113 = 1213.

Q5 (NCERT / DHSE March • 3 Marks): Given P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B), (ii) P(A' ∩ B'), (iii) P(A ∩ B').

• (i) P(A ∪ B) = 0.54 + 0.69 − 0.35 = 0.88.
• (ii) P(A' ∩ B') = P((A ∪ B)') = 1 − 0.88 = 0.12.
• (iii) P(A ∩ B') = P(A) − P(A ∩ B) = 0.54 − 0.35 = 0.19.

Q6 (Kerala HSE March • 3 Marks): In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both. If one student is selected at random, find the probability that the student opted for: (i) NCC or NSS, (ii) neither NCC nor NSS, (iii) NSS but not NCC.

P(NCC) = 30/60, P(NSS) = 32/60, P(NCC ∩ NSS) = 24/60.
• (i) P(NCC ∪ NSS) = 30 + 32 − 2460 = 3860 = 1930.
• (ii) P(neither) = 1 − 1930 = 1130.
• (iii) P(NSS − NCC) = 32 − 2460 = 860 = 215.

Q7 (Kerala HSE Say • 3 Marks): A committee of two persons is selected from two men and two women. What is the probability that the committee will have (a) no man, (b) one man, (c) two men?

Total ways = ⁴C₂ = 6.
• (a) No man (2 women): ²C₂⁶ = 16.
• (b) One man & one woman: ²C₁ × ²C₁6 = 46 = 23.
• (c) Two men: ²C₂6 = 16. (Check: 1/6 + 2/3 + 1/6 = 1!)

Q8 (NCERT / DHSE Model • 4 Marks): Two students Anil and Ashima appeared in an exam. P(Anil qualifies) = 0.05, P(Ashima qualifies) = 0.10, and P(both qualify) = 0.02. Find P that: (a) both will not qualify, (b) at least one will not qualify, (c) only one of them will qualify.

Let E = Anil qualifies, F = Ashima qualifies. P(E ∪ F) = 0.05 + 0.10 − 0.02 = 0.13.
• (a) Both not: P(E' ∩ F') = 1 − P(E ∪ F) = 1 − 0.13 = 0.87.
• (b) At least one not: P(E' ∪ F') = 1 − P(E ∩ F) = 1 − 0.02 = 0.98.
• (c) Only one: P(E) + P(F) − 2P(E ∩ F) = 0.05 + 0.10 − 2(0.02) = 0.11.

Q9 (Kerala HSE March • 4 Marks): Two dice are thrown. A: getting an even number on first die; B: getting an odd number on first die; C: getting sum ≤ 5. Describe (i) A', (ii) not B, (iii) A or B, (iv) A and B, (v) A and not C.

n(S) = 36.
• (i) A' = odd on first die = B (18 pairs).
• (ii) not B = even on first die = A (18 pairs).
• (iii) A or B = A ∪ B = S (all 36 pairs; exhaustive!).
• (iv) A and B = A ∩ B = ∅ (mutually exclusive!).
• C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}.
• (v) A ∩ C' = first die even AND sum > 5 = 14 pairs: all 18 in A except (2,1),(2,2),(2,3),(4,1).