Crack the Exam: One Topic at a Time

Theory → Formula maps → Solved question bank → Calculator drills → Self-test. Built exactly around your three-unit syllabus and the Casio fx-82MS.

🌫️

Unit 1 · Fuzzy Sets

Membership grades, A′, ∪, ∩, −, algebraic sum/product, proofs, I-V sets, fuzzy relations + interactive fuzzy calculator.

🎲

Unit 2 · Probability

Binomial, Poisson & Normal — which formula where, live graphs, all 13 question-bank problems solved with keystrokes.

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Unit 3 · Groups

Group axioms, Cayley tables and the exact 4-step proof-writing template used in all 5 model problems.

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Calculator Lab

nCr, e^x, x!, memory tricks, the Poisson ladder — save 10+ minutes in the exam hall.

📋

Cheat Sheet

Every formula on one screen for the final night of revision.

✅

Self-Test Quiz

10 rapid-fire questions across all units with instant feedback.

How to Use This Site (The 5-Step Loop)

Read the theory card for the topic — focus only on the boxed formulas.
Learn "which formula where" using the decision tables (most marks are lost here, not in arithmetic).
Attempt each question yourself first, then press Show Solution and compare line-by-line.
Redo the arithmetic using the fx-82MS keystroke sequences shown in green boxes until they're muscle memory.
Finish with the quiz. Below 8/10 → revisit the weak unit's question bank.

Syllabus Checklist

🌫️ Unit 1

Basic definitions of fuzzy sets
Operations: ′ ∩ ∪ −, + , ·
Image & inverse images
Interval-valued (I-V) fuzzy sets
Fuzzy relations

🎲 Unit 2

Binomial: PMF, mean, variance, properties, uses
Poisson: PMF, mean, variance, properties, applications
Normal: PDF, properties, importance + graphs

🧮 Unit 3

Algebraic structures with one binary operation
Group axioms & elementary properties
Cayley tables · roots of unity · R×R group · 2×2 matrix group

🔢 Exam Skills

nCr, e^x, x! keystrokes
"At least / at most" complement strategy
Proof-writing template (4 axioms)
Answer presentation to 4 decimals

⚠️ Typo alerts fixed in your question bank: Unit 1 Q2 uses B = {(x₃, 0.4)} (the "4" is a misprint — membership values must lie in [0,1]) and Q4 uses B = {(x₁,0.5),(x₂,0.3),(x₃,0.5)} (x₃ was duplicated).

🌫️ Unit 1 — Fuzzy Sets

A fuzzy set lets an element belong "partially". Master 6 operations and 3 proof patterns — that's the whole unit.

Definition

A fuzzy set \(A\) on a universe \(E\) is a set of ordered pairs \(A = \{(x, \mu_A(x)) : x \in E\}\) where the membership function \(\mu_A : E \to [0,1]\) gives the grade of membership of \(x\) in \(A\).

Read grades as: \(\mu_A(x)=1\) → full member · \(\mu_A(x)=0\) → not a member · \(0<\mu_A(x)<1\) → partial member.

The 6 Operations (Learn This Table = 80% of Unit 1)

Operation	Symbol	Membership formula	Mini-example (μ_A=0.2, μ_B=0.5)
Complement	\(A'\)	\(\mu_{A'}(x)=1-\mu_A(x)\)	1 − 0.2 = 0.8
Union	\(A\cup B\)	\(\max(\mu_A(x),\mu_B(x))\)	max(0.2, 0.5) = 0.5
Intersection	\(A\cap B\)	\(\min(\mu_A(x),\mu_B(x))\)	min(0.2, 0.5) = 0.2
Difference	\(A-B=A\cap B'\)	\(\min(\mu_A(x),\,1-\mu_B(x))\)	min(0.2, 0.5) = 0.2
Algebraic sum	\(A+B\)	\(\mu_A+\mu_B-\mu_A\mu_B\)	0.2+0.5−0.1 = 0.6
Algebraic product	\(A\cdot B\)	\(\mu_A(x)\,\mu_B(x)\)	0.2 × 0.5 = 0.1

🚫 Classic trap: \(A-B \neq\) ordinary subtraction of grades. It is defined as \(A\cap B'\): first complement B, then take min. Similarly "+" here is the algebraic sum, not plain addition.

Remaining Theory (Short-Answer Topics)

Image & Inverse Image

Let \(f:E\to F\) be an ordinary (crisp) function.

Image of A under f\(\mu_{f(A)}(y)=\sup\{\mu_A(x): f(x)=y\},\quad y\in F\)

Inverse image of B\(\mu_{f^{-1}(B)}(x)=\mu_B(f(x))\)

If several x map to the same y, take the largest grade; if none map, the grade is 0.

Interval-Valued (I-V) Fuzzy Sets

Membership is an interval: \(A=\{(x,[\mu_A^-(x),\mu_A^+(x)])\}\) with \([\mu_A^-,\mu_A^+]\subseteq[0,1]\).

Operations (endpoint-wise) \(A\cup B=[\max(\mu_A^-,\mu_B^-),\ \max(\mu_A^+,\mu_B^+)]\)
\(A\cap B=[\min(\mu_A^-,\mu_B^-),\ \min(\mu_A^+,\mu_B^+)]\)
\(A'=[1-\mu_A^+,\ 1-\mu_A^-]\)

Fuzzy Relations

A fuzzy relation \(R\) from \(X\) to \(Y\) is a fuzzy subset of \(X\times Y\): \(\mu_R(x,y)\in[0,1]\) = strength of relationship.

Max–Min composition\(\mu_{R\circ S}(x,z)=\max_{y}\ \min\big(\mu_R(x,y),\mu_S(y,z)\big)\)

Example: with R=[[0.2,0.5],[0.7,0.1]], S=[[0.4,0.8],[0.6,0.3]], \((R\circ S)_{11}=\max(\min(0.2,0.4),\min(0.5,0.6))=\max(0.2,0.5)=0.5\).

Proof Toolkit (for Q3-type theory questions)

\(\min(a,b)=\min(b,a)\) and \(\max(a,b)=\max(b,a)\) → commutativity.
\(1-\max(a,b)=\min(1-a,1-b)\) and \(1-\min(a,b)=\max(1-a,1-b)\) → De Morgan.
\(\max(a,\min(b,c))=\min(\max(a,b),\max(a,c))\) → distributivity (prove by assuming WLOG \(b\le c\)).

💡 Exam habit: in proofs always write "\(\forall x\in E\)" and end with "hence the equality holds for every x, so the sets are equal".

OperationsQ1. \(E=\{x_1,x_2,x_3,x_4\}\), \(A=\{(x_1,0.2),(x_2,1),(x_3,0.8),(x_4,0.5)\}\), \(B=\{(x_1,0.5),(x_2,0.1),(x_3,0.2),(x_4,0.3)\}\). Find \(A',\ B',\ A\cap B,\ A\cup B,\ A-B\).

Step 1 — Complements (subtract every grade from 1):

	x₁	x₂	x₃	x₄
A	0.2	1	0.8	0.5
B	0.5	0.1	0.2	0.3
A′	0.8	0	0.2	0.5
B′	0.5	0.9	0.8	0.7

Step 2 — min / max columns:

	x₁	x₂	x₃	x₄
A∩B = min	0.2	0.1	0.2	0.3
A∪B = max	0.5	1	0.8	0.5
A−B = min(A, B′)	0.2	0.9	0.8	0.5

A−B: e.g. at x₂ → min(1, 1−0.1) = min(1, 0.9) = 0.9 ✓

OperationsQ2. \(A=\{(x_1,0.6),(x_2,0.4),(x_3,0.1),(x_4,0)\}\), \(B=\{(x_1,0),(x_2,0.2),(x_3,0.4),(x_4,0.1)\}\). Find \(A',\ B',\ (A\cap B)',\ (A\cup B)',\ B-A\).

Step 1 — complements and min/max:

	x₁	x₂	x₃	x₄
A	0.6	0.4	0.1	0
B	0	0.2	0.4	0.1
A′	0.4	0.6	0.9	1
B′	1	0.8	0.6	0.9
A∩B	0	0.2	0.1	0
(A∩B)′	1	0.8	0.9	1
A∪B	0.6	0.4	0.4	0.1
(A∪B)′	0.4	0.6	0.6	0.9
B−A=min(B,A′)	0	0.2	0.4	0.1

Note (A∩B)′ = 1 − (A∩B) elementwise; here B−A happened to equal B itself because A is tiny where B is large.

ProofQ3. For fuzzy subsets A, B, C of E prove: (a) \(A\cap B=B\cap A\) (b) \(A\cup(B\cap C)=(A\cup B)\cap(A\cup C)\) (c) \((A\cup B)'=A'\cap B'\) (De Morgan).

(a) For every \(x\in E\): \(\mu_{A\cap B}(x)=\min(\mu_A(x),\mu_B(x))=\min(\mu_B(x),\mu_A(x))=\mu_{B\cap A}(x)\). Since min is symmetric, \(A\cap B=B\cap A\). ∎

(b) Write \(a=\mu_A(x), b=\mu_B(x), c=\mu_C(x)\). We must show \(\max(a,\min(b,c))=\min(\max(a,b),\max(a,c))\).
WLOG assume \(b\le c\) (the case \(c\le b\) is identical by symmetry). Then \(\min(b,c)=b\), so LHS \(=\max(a,b)\).
Also \(\max(a,b)\le\max(a,c)\), so RHS \(=\min(\max(a,b),\max(a,c))=\max(a,b)\).
LHS = RHS \(\forall x\in E\), hence \(A\cup(B\cap C)=(A\cup B)\cap(A\cup C)\). ∎

(c) For every \(x\in E\):
\(\mu_{(A\cup B)'}(x)=1-\mu_{A\cup B}(x)=1-\max(\mu_A(x),\mu_B(x))\)
\(=\min(1-\mu_A(x),\,1-\mu_B(x))\quad(\because 1-\max(a,b)=\min(1-a,1-b))\)
\(=\min(\mu_{A'}(x),\mu_{B'}(x))=\mu_{A'\cap B'}(x)\). ∎

Presentation tip: every fuzzy proof = "for each x, show the two membership VALUES are equal", using only min/max identities.

Algebraic +, ·Q4. \(A=\{(x_1,0.2),(x_2,1),(x_3,0.7)\}\), \(B=\{(x_1,0.5),(x_2,0.3),(x_3,0.5)\}\). Find \(A+B\) and \(A\cdot B\).

Algebraic sum \(\mu_{A+B}(x)=\mu_A(x)+\mu_B(x)-\mu_A(x)\mu_B(x)\):

x₁: \(0.2+0.5-(0.2)(0.5)=0.7-0.1=0.6\)
x₂: \(1+0.3-(1)(0.3)=1\)
x₃: \(0.7+0.5-(0.7)(0.5)=1.2-0.35=0.85\)

Algebraic product \(\mu_{A\cdot B}(x)=\mu_A(x)\mu_B(x)\):

x₁: \(0.2\times0.5=0.1\)
x₂: \(1\times0.3=0.3\)
x₃: \(0.7\times0.5=0.35\)

A+B = {(x₁,0.6),(x₂,1),(x₃,0.85)} · A·B = {(x₁,0.1),(x₂,0.3),(x₃,0.35)}

fx-82MS (sum, x₃): 0.7 + 0.5 − 0.7 × 0.5 = → 0.85

🧰 Fuzzy Operation Calculator — pre-loaded with Q1's data

Edit any grade (0–1) and press Compute. Use it to verify every hand answer.

	x₁	x₂	x₃	x₄	x₅
A
B

🎲 Unit 2 — Probability Distributions

The #1 skill: recognise the distribution from the story. Then it's one formula plus calculator work.

Recognition Table — memorise the keywords

Story smells like…	Keywords in the question	Use
Fixed number of repeated yes/no trials	"coin tossed n times", "n items produced", "40 MCQs guessed", "5 children"	Binomial
Counting events over time/space with an average rate	"per minute", "per page", "average number … is λ", "rare accidents"	Poisson
Binomial with n large AND p tiny	"p = 0.005, n = 400", "10 in 100000", "use Poisson approximation"	Poisson approx λ = np
Continuous measurement, bell-shaped	"heights", "marks", "normally distributed", μ and σ given	Normal

🧭 Formula Finder (tap through)

"At least / At most" Phrase Decoder

Phrase	Translate	Smart move
Exactly k	\(P(X=k)\)	direct formula
At most k	\(P(X\le k)\)	add terms 0…k
At least k	\(P(X\ge k)\)	\(1-P(X\le k-1)\) ← fewer terms!
More than k	\(P(X>k)\)	\(1-P(X\le k)\)
Between a and b	\(P(a\le X\le b)\)	add terms a…b
At least one	\(P(X\ge1)\)	\(1-P(X=0)\) ← always!

💡 Golden rule: whichever side has fewer terms — compute THAT side and subtract from 1 if needed.

Approximation Rules (when may I switch?)

Binomial → Poisson: allowed when \(n\) is large (\(n\ge 20\)) and \(p\) is small (\(p\le 0.05\)) with \(\lambda=np\) moderate. (Used in Q10–Q13 of your bank.)
Binomial → Normal: when \(n\) is large and \(p\) is NOT tiny (np and nq both ≥ 5): use \(\mu=np,\ \sigma=\sqrt{npq}\).
Poisson mean–variance check: if a problem states mean ≈ variance, think Poisson.

Binomial

PMF\(P(X=x)=\binom{n}{x}p^x q^{\,n-x},\ x=0,1,\dots,n\)

Mean / Variance / SD\(\mu=np,\quad \sigma^2=npq,\quad \sigma=\sqrt{npq}\)

Properties

Fixed n independent trials, two outcomes, constant p, q = 1−p.
Mean > Variance always (since q < 1).
Symmetric when p = ½; \(\sum P(x)=(q+p)^n=1\).

Uses

Coins/dice, defective items in a batch, guessing MCQs, boys/girls in families, hit-or-miss trials.

Poisson

PMF\(P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!},\ x=0,1,2,\dots\)

Mean & Variance\(\mu=\lambda,\qquad \sigma^2=\lambda\)

Properties

Mean = Variance = λ (signature property!).
Events independent; two events can't occur at the same instant.
Limit of Binomial: n→∞, p→0, np=λ fixed.
Sums of independent Poissons are Poisson (λ's add).

Applications

Phone calls per minute, printing mistakes per page, accidents per year, rare defects, arrivals at a counter.

Normal

PDF\(f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-(x-\mu)^2/2\sigma^2}\)

Standardise\(Z=\dfrac{X-\mu}{\sigma},\ Z\sim N(0,1)\)

Properties

Bell-shaped, symmetric about x = μ.
Mean = Median = Mode = μ.
Total area = 1; curve never touches the axis.
68-95-99.7 rule within μ±σ, μ±2σ, μ±3σ.

Importance

Central Limit Theorem; approximates Binomial/Poisson for large parameters; models most natural measurements (heights, errors, marks).

⚠️ Mean/variance algebra (Q2 pattern): given np and npq, divide: \(\dfrac{npq}{np}=q\) → then \(p=1-q\) → then \(n=\dfrac{np}{p}\). Three divisions, done.

Binomial PMF — B(n,p)

n (trials)12

p (success)0.50

Move p away from 0.5 → the bars skew. This is Q6's shape (p=0.1).

Poisson PMF — P(λ)

λ (rate)2.4

Bars peak near x ≈ λ. Q7 uses λ = 2.35 — try it!

Normal PDF — N(μ, σ²)

μ (mean)50

σ (std dev)10

Larger σ → flatter, wider bell; μ only slides the curve left/right. Shaded band = μ ± σ (≈ 68% of area).

🧰 Exam Solver — verify every answer you compute by hand

Binomial Poisson

n p

—

Workflow: solve by hand first → type same n, p, x here → if your 4-decimal answer matches, your keystrokes are correct.

💡 Study method: hide the solution, write your own, then compare. Every solution below ends with the exact fx-82MS key sequence.

BinomialQ1. A coin is tossed 4 times. Find P( (1) exactly two heads (2) at least two heads (3) at most two heads ).

Here \(n=4,\ p=q=\tfrac12\), so \(P(x)=\binom{4}{x}\left(\tfrac12\right)^4=\dfrac{\binom{4}{x}}{16}\).

(1) \(P(2)=\dfrac{\binom{4}{2}}{16}=\dfrac{6}{16}=\tfrac{3}{8}=0.375\)

(2) \(P(X\ge2)=\dfrac{\binom{4}{2}+\binom{4}{3}+\binom{4}{4}}{16}=\dfrac{6+4+1}{16}=\dfrac{11}{16}=0.6875\)

(3) \(P(X\le2)=\dfrac{\binom{4}{0}+\binom{4}{1}+\binom{4}{2}}{16}=\dfrac{1+4+6}{16}=\dfrac{11}{16}=0.6875\)

4 SHIFT+÷(nCr) 2 × 0.5 ^ 4 = → 0.375 · then × 16 = shows numerator 6 → 6/16 = 3/8

Answers: 3/8 · 11/16 · 11/16

BinomialFormula algebraQ2. Mean and SD of a binomial distribution are 5 and 2. Determine the distribution.

\(np=5\) and \(\sqrt{npq}=2 \Rightarrow npq=4\).

Divide: \(q=\dfrac{npq}{np}=\dfrac{4}{5}=0.8 \Rightarrow p=1-q=0.2\).

Then \(n=\dfrac{np}{p}=\dfrac{5}{0.2}=25\).

\(P(X=x)=\dbinom{25}{x}\left(\tfrac15\right)^x\left(\tfrac45\right)^{25-x},\ x=0,1,\dots,25\) — i.e. B(25, 0.2).

BinomialQ3. A die is thrown 6 times. Getting an odd number is a success. Find P( (1) five successes (2) at least five successes ).

Odd numbers on a die: {1,3,5} ⇒ \(p=\tfrac36=\tfrac12,\ n=6\). \(P(x)=\binom{6}{x}/64\).

(1) \(P(5)=\dfrac{\binom{6}{5}}{64}=\dfrac{6}{64}=\dfrac{3}{32}\approx0.0938\)

(2) \(P(X\ge5)=P(5)+P(6)=\dfrac{6+1}{64}=\dfrac{7}{64}\approx0.1094\)

6 SHIFT+÷ 5 = → 6 (that's C(6,5)); ÷ 64 = → 0.09375

Answers: 3/32 · 7/64

BinomialQ4. An MCQ test has 40 questions with 5 options each. A student guesses all. Find the mean and SD of the number correct.

\(n=40,\ p=\tfrac15=0.2,\ q=0.8\).

Mean \(\mu=np=40\times0.2=8\)

Variance \(=npq=40\times0.2\times0.8=6.4 \Rightarrow \sigma=\sqrt{6.4}\approx2.5298\)

40 × 0.2 × 0.8 = → 6.4 then √ Ans = → 2.5298

Mean = 8, SD ≈ 2.53 correct answers.

BinomialExpected countQ5. Out of 800 families with 5 children each, how many are expected to have (1) 3 boys (2) 5 girls (3) 2 or 3 boys (4) at least one boy?

\(n=5,\ p=\tfrac12 \Rightarrow P(x)=\binom{5}{x}/32\). Expected families \(=800\times P\).

(1) \(P(3\text{ boys})=\dfrac{\binom{5}{3}}{32}=\dfrac{10}{32}\Rightarrow 800\times\dfrac{10}{32}=250\)

(2) 5 girls = 0 boys: \(P(0)=\dfrac{1}{32}\Rightarrow 800/32=25\)

(3) \(P(2)+P(3)=\dfrac{10+10}{32}=\dfrac{20}{32}\Rightarrow 500\)

(4) \(P(X\ge1)=1-P(0)=1-\dfrac{1}{32}=\dfrac{31}{32}\Rightarrow 800\times\dfrac{31}{32}=775\)

250 · 25 · 500 · 775 families. Trick: all denominators are 32, so expectations are just 800×numerator/32 = 25×numerator.

BinomialQ6. P(defective) = 1/10. Of 12 items, find P( (1) exactly one defective (2) at least two defective (3) none defective ).

\(n=12,\ p=0.1,\ q=0.9\).

(1) \(P(1)=\binom{12}{1}(0.1)(0.9)^{11}=12(0.1)(0.31381)\approx0.3766\)

(3) \(P(0)=(0.9)^{12}\approx0.2824\)

(2) \(P(X\ge2)=1-P(0)-P(1)=1-0.2824-0.3766\approx0.3410\)

P(1): 12 SHIFT+÷ 1 × 0.1 × 0.9 ^ 11 = → 0.3766
P(0): 0.9 ^ 12 = → 0.2824 · P(≥2): 1 − Ans − 0.3766 = → 0.341

0.3766 · 0.3410 · 0.2824

PoissonQ7. Average phone calls per minute = 2.35. Find P(at most 2 calls in a minute).

Rate per minute given → Poisson with \(\lambda=2.35\).

\(P(X\le2)=e^{-2.35}\left(1+2.35+\dfrac{2.35^2}{2!}\right)=e^{-2.35}(1+2.35+2.7613)=e^{-2.35}(6.1113)\)

\(=0.095369\times6.1113\approx0.5829\)

e^(−2.35): SHIFT+ln(eˣ) (−) 2.35 = → 0.0954 · × 6.1113 = → 0.5829

P(X ≤ 2) ≈ 0.5829

PoissonFind λ firstQ8. P(X=1) = P(X=2). Find (1) mean (2) P(X=4) (3) P(X≥1) (4) P(1 < X < 4).

\(\dfrac{e^{-\lambda}\lambda^1}{1!}=\dfrac{e^{-\lambda}\lambda^2}{2!}\Rightarrow \lambda=\dfrac{\lambda^2}{2}\Rightarrow \lambda^2-2\lambda=0\Rightarrow \lambda=2\) (λ=0 rejected).

(1) Mean \(=\lambda=2\)

(2) \(P(4)=\dfrac{e^{-2}\,2^4}{4!}=\dfrac{16e^{-2}}{24}=\tfrac23e^{-2}\approx0.0902\)

(3) \(P(X\ge1)=1-P(0)=1-e^{-2}\approx1-0.1353=0.8647\)

(4) \(P(1<X<4)=P(2)+P(3)=e^{-2}\left(\dfrac{4}{2}+\dfrac{8}{6}\right)=e^{-2}\cdot\dfrac{10}{3}\approx0.4511\)

e^(−2): SHIFT+ln (−) 2 = → 0.1353 · store: SHIFT+RCL(STO) (-)(A) · P(4): RCL A × 2 ^ 4 ÷ 4 SHIFT+x⁻¹(x!) = → 0.0902

λ = 2 · 0.0902 · 0.8647 · 0.4511

PoissonFind λ firstQ9. Given \(\tfrac32 P(X=1)=P(X=3)\), find (i) P(X≥1) (ii) P(X≤3) (iii) P(2≤X≤5).

\(\tfrac32\,e^{-\lambda}\lambda=\dfrac{e^{-\lambda}\lambda^3}{3!}\Rightarrow \tfrac32\lambda=\dfrac{\lambda^3}{6}\Rightarrow 9\lambda=\lambda^3\Rightarrow \lambda^2=9\Rightarrow \lambda=3\).

(i) \(P(X\ge1)=1-e^{-3}=1-0.0498=0.9502\)

(ii) \(P(X\le3)=e^{-3}\left(1+3+\dfrac{9}{2}+\dfrac{27}{6}\right)=13e^{-3}\approx0.6472\)

(iii) \(P(2\le X\le5)=e^{-3}\left(\dfrac{9}{2}+\dfrac{27}{6}+\dfrac{81}{24}+\dfrac{243}{120}\right)=e^{-3}(4.5+4.5+3.375+2.025)=14.4\,e^{-3}\approx0.7169\)

SHIFT+ln (−) 3 = → 0.0498 · × 14.4 = → 0.7169

0.9502 · 0.6472 · 0.7169

Poisson approxQ10. P(a page has ≥1 error) = 0.005. In a 400-page novel find P( (1) exactly one page with errors (2) at most three pages with errors ).

n = 400 large, p = 0.005 tiny → Poisson approx with \(\lambda=np=400\times0.005=2\).

(1) \(P(1)=\dfrac{e^{-2}\,2^1}{1!}=2e^{-2}\approx0.2707\)

(2) \(P(X\le3)=e^{-2}\left(1+2+\dfrac{4}{2}+\dfrac{8}{6}\right)=e^{-2}\times\dfrac{19}{3}\approx0.1353\times6.3333=0.8571\)

0.2707 · 0.8571 — Note the tell-tale signs: "large n, tiny p" → immediately write λ = np.

Poisson approxQ11. 4000 people insured; on average 10 in 100000 suffer the injury yearly. Find P(more than 3 claims in a year).

\(p=\dfrac{10}{100000}=0.0001,\ n=4000 \Rightarrow \lambda=np=0.4\).

\(P(X>3)=1-P(X\le3)=1-e^{-0.4}\left(1+0.4+\dfrac{0.4^2}{2!}+\dfrac{0.4^3}{3!}\right)\)

\(=1-e^{-0.4}(1+0.4+0.08+0.010667)=1-(0.670320)(1.490667)\)

\(=1-0.999224\approx0.0008\)

SHIFT+ln (−) 0.4 = → 0.6703 · × 1.490667 = → 0.9992 · 1 − Ans = → 0.0008

≈ 0.000776 ≈ 0.0008 (extremely unlikely — as expected!)

PoissonQ12. A 585-page book has 43 errors, randomly distributed. Find P(10 randomly selected pages are error-free).

Rate per page: \(\lambda=\dfrac{43}{585}\approx0.0735\) errors/page.

P(one page error-free) \(=P(0)=e^{-0.0735}\approx0.9291\).

Pages independent ⇒ P(all 10 free) \(=(e^{-0.0735})^{10}=e^{-0.735}\approx0.4795\).

43 ÷ 585 = → 0.0735 · × 10 = → 0.7350 · SHIFT+ln (−) Ans = → 0.4795 (uses e^(−10λ) directly!)

≈ 0.4795

Poisson approxQ13. Six coins are tossed 6400 times. Using Poisson, find the approximate probability of getting six heads exactly 10 times.

P(six heads in one throw of 6 coins) \(=p=\left(\tfrac12\right)^6=\dfrac{1}{64}\).

\(n=6400\) trials ⇒ \(\lambda=np=\dfrac{6400}{64}=100\).

\(P(X=10)=\dfrac{e^{-100}\,100^{10}}{10!}\approx1.03\times10^{-30}\approx0\).

Effectively zero — with λ = 100 you'd expect ~100 occurrences, so exactly 10 is astronomically rare. (If your paper intended "100 times", the same method gives e⁻¹⁰⁰·100¹⁰⁰/100! ≈ 0.0399.)

🧮 Unit 3 — Algebraic Structures & Morphisms

Every exam answer here is the same 4-step skeleton. Learn the template once; reuse it forever.

Standard Sets

Symbol	Set
\(\mathbb{N}\)	{1, 2, 3, …}
\(W\)	{0, 1, 2, 3, …}
\(\mathbb{Z}\)	{…, −2, −1, 0, 1, 2, …}
\(\mathbb{Q}\)	{p/q : p ∈ ℤ, q ∈ ℕ}
\(\mathbb{R}\)	all real numbers
\(GL_2\)	2×2 real matrices with ad − bc ≠ 0
Roots of unity	{x ∈ ℂ : xⁿ = 1}

Modular Arithmetic (read carefully — used in Cayley tables)

\(3+_5 4=2\) because \(3+4=7\equiv2\pmod 5\)
\(3\times_4 7=1\) because \(21\equiv1\pmod 4\)
\(a\equiv b \pmod m \iff m\mid(a-b)\)

Binary Operation & Algebraic Structure

A binary operation * on G combines two elements to give one result. A non-empty set G with one or more binary operations is an algebraic structure, e.g. \((\mathbb{Z},+),\ (\mathbb{R},+,\times)\).

⭐ Definition of a Group — THE four axioms

\((G,*)\) is a group if for all elements of G:

G1 · Closure\(\forall\, a,b\in G \Rightarrow a*b\in G\)

G2 · Associativity\((a*b)*c=a*(b*c)\quad \forall\,a,b,c\in G\)

G3 · Identity\(\exists\, e\in G:\ a*e=a=e*a\quad\forall a\in G\)

G4 · Inverse\(\forall a\in G,\ \exists\, a^{-1}\in G:\ a*a^{-1}=e=a^{-1}*a\)

Add G5 · Commutativity \(a*b=b*a\) and it's an abelian (commutative) group.

Elementary Properties (quote these in theory questions)

Property	Statement
Uniqueness	Identity e and each inverse a⁻¹ are unique.
Double inverse	\((a^{-1})^{-1}=a\)
Reversal law	\((ab)^{-1}=b^{-1}a^{-1}\) (note the ORDER flips!)
Left cancellation	\(ab=ac \Rightarrow b=c\)
Right cancellation	\(ba=ca \Rightarrow b=c\)

✍️ The Universal Proof Template (copy this structure in the exam)

Claim: (G, *) is a group.
(i) Closure: Let a, b ∈ G. Then a*b = … ∈ G (∵ reason).
(ii) Associativity: Let a, b, c ∈ G. LHS = a*(b*c) = … ; RHS = (a*b)*c = … . LHS = RHS.
(iii) Identity: Suppose a*e = a. Solving gives e = … . Check e*a = a too. e ∈ G. ✓
(iv) Inverse: Suppose a*a′ = e. Solving gives a′ = … . Check a′*a = e. a′ ∈ G. ✓
Conclusion: Hence from (i)–(iv), (G, *) is a group. [+ (v) a*b = b*a ⇒ abelian.]

💡 How to FIND the identity: set up the equation \(a*e=a\), treat e as the unknown, solve. Always verify from the other side (\(e*a=a\)) — in non-commutative operations both sides must be checked.

💡 How to FIND the inverse: set \(a*a'=e\) using the e you just found, solve for a′, then verify \(a'*a=e\) and confirm \(a'\in G\) (e.g. needs a ≠ 0 in Problem 4).

🚫 Mark-losers: (1) forgetting to state that the identity/inverse actually lies IN G; (2) checking the operation from one side only; (3) in Cayley-table questions, not explicitly reading closure off the table.

Cayley Table Facts (fast checks for finite groups)

Closure: every cell entry belongs to the set.
Identity: the row/column that repeats the header order.
Inverses: e appears exactly once in every row & column; pairs are symmetric.
Commutativity: table is symmetric about the main diagonal.

Cayley tableProblem 1. Draw the composition table for x * y = x on S = {a, b, c, d}.

Rule: row element is always the answer (x * y = x), so every entry in row x equals x:

*	a	b	c	d
a	a	a	a	a
b	b	b	b	b
c	c	c	c	c
d	d	d	d	d

Each row is constant — that's the whole trick for "x * y = x".

Finite groupProblem 2. Show that the square roots of unity form a group under multiplication.

\(x^2=1 \Rightarrow x=\pm1\), so \(G=\{1,-1\}\).

×	1	−1
1	1	−1
−1	−1	1

(i) Closure: every entry in the table ∈ G. ✓

(ii) Associativity: usual multiplication of reals is associative. ✓

(iii) Identity: \(1\times1=1,\ (-1)\times1=-1\) ⇒ \(e=1\in G\). ✓

(iv) Inverse: \(1\times1=1\Rightarrow1^{-1}=1\); \((-1)(-1)=1\Rightarrow(-1)^{-1}=-1\). Both ∈ G. ✓

Hence from (i)–(iv), (G, ×) is a group. ∎

CounterexampleProblem 3. Show x * y = yˣ on ℕ is neither commutative nor associative.

Not commutative: take 2, 3 ∈ ℕ: \(2*3=3^2=9\) but \(3*2=2^3=8\). Since \(2*3\ne3*2\), * is not commutative.

Not associative: take 2, 3, 4 ∈ ℕ:
LHS \(=2*(3*4)=2*(4^3)=2*64=64^2=4096\) …(1)
RHS \(=(2*3)*4=9*4=4^9=262144\) …(2)

From (1), (2): \(2*(3*4)\ne(2*3)*4\). Not associative. ∎

To DISPROVE a property, ONE counterexample with full working earns the marks — pick small numbers like 2, 3, 4.

Infinite groupExam favouriteProblem 4. G = {(a,b) : a,b ∈ ℝ, a ≠ 0} with (a,b)*(c,d) = (ac, bc+d). Show (G,*) is a group.

(i) Closure: \((a,b)*(c,d)=(ac,bc+d)\in G\) since \(a,b,c,d\in\mathbb{R}\Rightarrow ac,\ bc+d\in\mathbb{R}\) (and ac ≠ 0 as a,c ≠ 0). ✓

(ii) Associativity:
LHS \(=(a,b)*\big((c,d)*(e,f)\big)=(a,b)*(ce,de+f)=(ace,\ bce+de+f)\) …(1)
RHS \(=\big((a,b)*(c,d)\big)*(e,f)=(ac,bc+d)*(e,f)=(ace,\ (bc+d)e+f)=(ace,\ bce+de+f)\) …(2)
LHS = RHS ⇒ associative. ✓

(iii) Identity: Let \((e_1,e_2)\) satisfy \((a,b)*(e_1,e_2)=(a,b)\):
\((ae_1,\ be_1+e_2)=(a,b)\Rightarrow ae_1=a\Rightarrow e_1=1;\ be_1+e_2=b\Rightarrow b+e_2=b\Rightarrow e_2=0\).
Check other side: \((1,0)*(a,b)=(1\cdot a,\ 0\cdot a+b)=(a,b)\) ✓. So \(e=(1,0)\in G\).

(iv) Inverse: Let \((a,b)*(a',b')=(1,0)\):
\((aa',\ ba'+b')=(1,0)\Rightarrow aa'=1\Rightarrow a'=\dfrac1a;\ \dfrac{b}{a}+b'=0\Rightarrow b'=-\dfrac{b}{a}\).
Check other side: \(\left(\tfrac1a,-\tfrac{b}{a}\right)*(a,b)=\left(\tfrac1a\cdot a,\ -\tfrac{b}{a}\cdot a+b\right)=(1,0)\) ✓.
So \((a,b)^{-1}=\left(\dfrac1a,-\dfrac{b}{a}\right)\in G\) (needs a ≠ 0 — that's why G excludes a = 0).

From (i)–(iv), (G, *) is a group. ∎ (Not abelian: (a,b)*(c,d) ≠ (c,d)*(a,b) in general.)

Matrix groupProblem 5. \(G=\left\{\begin{bmatrix}a&b\\-b&a\end{bmatrix}:a,b\in\mathbb{R}\right\}\). Show G is an abelian group under matrix addition.

Let \(A=\begin{bmatrix}a&b\\-b&a\end{bmatrix}, B=\begin{bmatrix}c&d\\-d&c\end{bmatrix}, C=\begin{bmatrix}e&f\\-f&e\end{bmatrix}\in G\).

(i) Closure: \(A+B=\begin{bmatrix}a+c&b+d\\-(b+d)&a+c\end{bmatrix}\) — same pattern (top-right = −(bottom-left), equal diagonals) ⇒ ∈ G. ✓

(ii) Associativity: matrix addition is entry-wise addition of reals, which is associative: \(A+(B+C)=(A+B)+C\) (verify by adding entries; e.g. top-left \(a+(c+e)=(a+c)+e\)). ✓

(iii) Identity: Need \(E\) with \(A+E=A\): \(\begin{bmatrix}a+e_1&b+e_2\\-b+e_3&a+e_4\end{bmatrix}=\begin{bmatrix}a&b\\-b&a\end{bmatrix}\Rightarrow e_1=e_2=e_3=e_4=0\).
\(E=\begin{bmatrix}0&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\-0&0\end{bmatrix}\in G\) and \(A+E=A=E+A\). ✓

(iv) Inverse: Need \(A'\) with \(A+A'=E\): entries give \(a_1=-a,\ a_2=-b,\ a_3=b,\ a_4=-a\).
\(A'=\begin{bmatrix}-a&-b\\b&-a\end{bmatrix}=\begin{bmatrix}-a&-b\\-(-b)&-a\end{bmatrix}\in G\) — it has the required pattern. \(A+A'=E=A'+A\). ✓

(v) Commutativity: \(A+B=\begin{bmatrix}a+c&b+d\\-(b+d)&a+c\end{bmatrix}=\begin{bmatrix}c+a&d+b\\-(d+b)&c+a\end{bmatrix}=B+A\) (real addition commutes). ✓

From (i)–(v), G is an abelian group under addition. ∎ Key exam phrase: "the result has the same pattern, hence it belongs to G."

🔢 fx-82MS Calculator Lab

Keystroke-level tricks mapped to this exact syllabus. Practise each recipe 3 times before the exam.

1 · Key Map (where everything hides)

You need	Press	Yellow label location
nCr (combinations)	`n` `SHIFT` `÷` `r` `=`	above the ÷ key
nPr	`SHIFT` `×`	above the × key
Factorial x!	`n` `SHIFT` `x⁻¹`	above the x⁻¹ key
e^x	`SHIFT` `ln`	above the ln key
Power a^b	`a` `^` `b`	dedicated ^ key
Negative number	`(−)` then digits	the (−) key, NOT the minus key
Store to memory A	`SHIFT` `RCL`(STO) `(-)`(A)	STO above RCL; red A on (−) key
Recall memory A	`RCL` `(-)`(A)	—
Add display into M	`M+`	recall with `RCL` `M+`(M)
Reuse last answer	`Ans` (auto-inserted if you press an operator first)	—
Fix 4 decimals	`MODE`×4 → `1`(Fix) → `4`	—
Full reset (before exam!)	`SHIFT` `MODE`(CLR) `3`(All) `=`	CLR above MODE

2 · Exam Recipes

Recipe A — Single binomial term

Goal: \(\binom{12}{1}(0.1)(0.9)^{11}\)

12 SHIFT+÷ 1 × 0.1 × 0.9 ^ 11 = → 0.3766

💡 Type the whole PMF left-to-right exactly as written — the fx-82MS follows BODMAS correctly.

Recipe B — "At least" via complement (Q6 pattern)

P(0): 0.9 ^ 12 = then SHIFT+RCL (-)(A)
P(1): 12 SHIFT+÷ 1 × 0.1 × 0.9 ^ 11 = then SHIFT+RCL °′″(B)
Final: 1 − RCL (-)(A) − RCL °′″(B) = → 0.341

💡 Storing each term in A, B, C beats retyping and prevents bracket errors.

Recipe C — Store e^(−λ) once (Q8–Q11 pattern)

SHIFT+ln (−) 2 = → 0.1353 then SHIFT+RCL (-)(A)
P(4) = e⁻²·2⁴/4! : RCL A × 2 ^ 4 ÷ ( 4 SHIFT+x⁻¹ ) = → 0.0902

Recipe D — ⭐ The Poisson Ladder (fastest trick in this syllabus)

Since \(\dfrac{P(x+1)}{P(x)}=\dfrac{\lambda}{x+1}\), each term = previous × λ ÷ (x+1). With Ans-chaining you never retype:

λ = 2.35:
SHIFT+ln (−) 2.35 = → 0.0954 = P(0)
× 2.35 ÷ 1 = → 0.2241 = P(1)
× 2.35 ÷ 2 = → 0.2633 = P(2)
…keep going for P(3), P(4)…

💡 Sum as you go: press M+ after each term, then RCL M+(M) gives P(X≤k) instantly. (Binomial version: multiply by (n−x)·p/((x+1)·q).)

Recipe E — Mean & SD in one line

n=40, p=0.2: 40 × 0.2 = → mean 8
× 0.8 = → variance 6.4 (Ans keeps the chain)
√ Ans = → SD 2.5298

Recipe F — Fuzzy complements & algebraic sum

Complement of 0.2: 1 − 0.2 = → 0.8 (chain: 1 − Ans for the next one!)
Alg. sum: 0.7 + 0.5 − 0.7 × 0.5 = → 0.85

3 · Traps That Cost Marks

🚫 Negative exponent: e^(−2.35) must use the (−) key — using the subtraction minus gives a syntax error or wrong sign.

🚫 Factorial of an expression: (2λ)! is NOT 2×λ! — always bracket before x!.

🚫 q = 1 − p: write it as the FIRST line of every binomial solution; mixing p and q mid-calculation is the #1 arithmetic error.

🚫 Rounding: keep full calculator precision till the final line, then round to 4 decimals. Never round intermediate terms.

⚠️ Before the exam: SHIFT → CLR → 3(All) → = to wipe memories; set Fix 4; verify with a known value like 5 nCr 2 = 10.

📋 One-Screen Cheat Sheet

Everything the exam can ask, condensed. (Use your browser's print for a paper copy.)

Unit 1

Fuzzy Operations @ x

\(A': 1-\mu_A\)
\(A\cup B: \max\)
\(A\cap B: \min\)
\(A-B: \min(\mu_A,1-\mu_B)\)
\(A+B: \mu_A+\mu_B-\mu_A\mu_B\)
\(A\cdot B: \mu_A\mu_B\)

Proof identities

\(1-\max(a,b)=\min(1-a,1-b)\)
\(\max(a,\min(b,c))=\min(\max(a,b),\max(a,c))\)

Unit 2

Binomial B(n,p)

\(P(x)=\binom{n}{x}p^xq^{n-x};\ \mu=np;\ \sigma^2=npq\)

Poisson P(λ)

\(P(x)=\frac{e^{-\lambda}\lambda^x}{x!};\ \mu=\sigma^2=\lambda\)

Normal N(μ,σ²)

\(f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2};\ Z=\frac{X-\mu}{\sigma}\)

n large + p small ⇒ Poisson, λ=np
"≥1" = 1−P(0) · ladder: P(x+1)=P(x)·λ/(x+1)

Unit 3

Group = 4 axioms

Closure: a*b ∈ G
Associativity: (a*b)*c = a*(b*c)
Identity: a*e = a = e*a
Inverse: a*a⁻¹ = e (a⁻¹ ∈ G)
+Commutativity ⇒ abelian

Key results

\((a*b)^{-1}=b^{-1}*a^{-1}\)
Cancellation: a*b=a*c ⇒ b=c
ℝ×ℝ group: e=(1,0), (a,b)⁻¹=(1/a, −b/a)
Matrix group: E = zero matrix, inverse = negate entries

✅ Self-Test Quiz

10 questions · all three units · instant feedback. Target: 8+/10.

Score: 0 / 10 answered correctly