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Mathematics and Statistics Part 2 गणित व संख्याशास्त्र भाग 2

Class 12 (Science, Arts) · 8 chapters · 8 with notes

📄 Maharashtra Board 12th (HSC) Question Papers 2008–2026 – download PDF

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Textbook · free PDF, ebalbharati.in

Chapters

  1. 1DifferentiationNotes
  2. 2Applications of DerivativesNotes
  3. 3Indefinite IntegrationNotes
  4. 4Definite IntegrationNotes
  5. 5Application of Definite IntegrationNotes
  6. 6Differential EquationsNotes
  7. 7Probability DistributionsNotes
  8. 8Binomial DistributionNotes

Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 1

Differentiation

Quick summary

This chapter extends derivatives to more functions. It covers the derivative of composite functions (chain rule), inverse functions, inverse trigonometric functions, logarithmic differentiation, implicit functions, parametric functions, and higher order derivatives.

Key points

  • Chain rule: if y = f(u) and u = g(x), then dy/dx = (dy/du)(du/dx).
  • Derivative of an inverse function: dx/dy = 1/(dy/dx), provided dy/dx ≠ 0.
  • Logarithmic differentiation is used for functions like xˣ, [f(x)]^g(x), or long products/quotients: take log on both sides first.
  • Implicit function: f(x, y) = 0. Differentiate both sides with respect to x, treating y as a function of x, then solve for dy/dx.
  • Parametric functions x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt).
  • Second derivative d²y/dx² is the derivative of dy/dx. For parametric functions, d²y/dx² = d/dt(dy/dx) ÷ (dx/dt).
  • Substitution (e.g. x = sin θ, x = tan θ) often simplifies derivatives of inverse trigonometric expressions.

Important terms

  • Composite function – function of a function, f(g(x)).
  • Implicit function – y not expressed directly in terms of x.
  • Parameter – a third variable t through which x and y are related.

Formulas

FunctionDerivative
sin⁻¹x1/√(1 − x²)
cos⁻¹x−1/√(1 − x²)
tan⁻¹x1/(1 + x²)
cot⁻¹x−1/(1 + x²)
sec⁻¹x1/(x√(x² − 1))
aˣaˣ log a
log x1/x
eˣeˣ
xⁿn xⁿ⁻¹
Product rule(uv)' = u v' + v u'
Quotient rule(u/v)' = (v u' − u v')/v²

5 important questions

Q1. Differentiate y = sin(x² + 5) with respect to x.

Let u = x² + 5. dy/dx = cos u × 2x = 2x cos(x² + 5).

Q2. Find dy/dx if y = xˣ.

log y = x log x. Differentiating: (1/y) dy/dx = x(1/x) + log x = 1 + log x. So dy/dx = xˣ (1 + log x).

Q3. Find dy/dx if x² + y² + xy = 7.

2x + 2y (dy/dx) + x (dy/dx) + y = 0. So dy/dx (2y + x) = −(2x + y), dy/dx = −(2x + y)/(x + 2y).

Q4. Find dy/dx if x = a cos³t and y = a sin³t.

dx/dt = −3a cos²t sin t; dy/dt = 3a sin²t cos t. dy/dx = (3a sin²t cos t)/(−3a cos²t sin t) = −tan t.

Q5. Differentiate y = tan⁻¹(2x/(1 − x²)) with respect to x (|x| < 1).

Put x = tan θ: 2x/(1 − x²) = tan 2θ, so y = 2θ = 2 tan⁻¹x. dy/dx = 2/(1 + x²).

Exam tip

Identify the type first (composite, implicit, parametric, log), then apply the right method. Learn the six inverse trig derivatives by heart.

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Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 2

Applications of Derivatives

Quick summary

Derivatives measure rates of change. This chapter uses them to find tangents and normals to curves, rates of change, approximations, Rolle's theorem and the Lagrange mean value theorem, increasing and decreasing functions, and maxima and minima.

Key points

  • Slope of the tangent at (x₁, y₁) is m = (dy/dx) at that point. Tangent: y − y₁ = m(x − x₁).
  • Slope of the normal = −1/m. Normal: y − y₁ = (−1/m)(x − x₁).
  • dy/dt gives the rate of change of y with time; related rates use the chain rule.
  • Approximation: f(a + h) ≈ f(a) + h f'(a), for small h.
  • Rolle's theorem: if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), there is c in (a, b) with f'(c) = 0.
  • Lagrange's mean value theorem (LMVT): there is c in (a, b) with f'(c) = (f(b) − f(a))/(b − a).
  • f is increasing where f'(x) > 0 and decreasing where f'(x) < 0.
  • Critical points: f'(x) = 0. Second derivative test: f''(c) < 0 gives maximum, f''(c) > 0 gives minimum.
  • First derivative test: if f' changes from + to − at c, maximum; from − to +, minimum.

Important terms

  • Tangent – line touching a curve at a point.
  • Normal – line perpendicular to the tangent at the point of contact.
  • Marginal cost/revenue – derivative of cost/revenue with respect to quantity (economics use).
  • Local maximum – value greater than at nearby points.
  • Stationary point – point where f'(x) = 0.

Formulas

UseFormula
Tangenty − y₁ = f'(x₁)(x − x₁)
Normaly − y₁ = −(1/f'(x₁))(x − x₁)
Approximationf(a + h) ≈ f(a) + h f'(a)
LMVTf'(c) = (f(b) − f(a))/(b − a)
Max/min testf'(c) = 0; f''(c) < 0 max, f''(c) > 0 min

5 important questions

Q1. Find the equations of the tangent and normal to y = x² + 1 at (1, 2).

dy/dx = 2x = 2 at x = 1. Tangent: y − 2 = 2(x − 1) ⇒ 2x − y = 0. Normal: y − 2 = −½(x − 1) ⇒ x + 2y − 5 = 0.

Q2. The radius of a circle increases at 0.5 cm/s. Find the rate of increase of its area when r = 4 cm.

A = π r². dA/dt = 2π r (dr/dt) = 2π × 4 × 0.5 = 4π cm²/s.

Q3. Find the approximate value of √(25.1).

f(x) = √x, a = 25, h = 0.1. f'(x) = 1/(2√x) = 1/10. √25.1 ≈ 5 + 0.1 × 0.1 = 5.01.

Q4. Find the maximum and minimum values of f(x) = x³ − 6x² + 9x + 1.

f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3) = 0 ⇒ x = 1, 3. f''(x) = 6x − 12. f''(1) = −6 < 0: maximum f(1) = 1 − 6 + 9 + 1 = 5. f''(3) = 6 > 0: minimum f(3) = 27 − 54 + 27 + 1 = 1.

Q5. Verify Rolle's theorem for f(x) = x² − 4x + 3 on [1, 3].

f is a polynomial, so continuous and differentiable. f(1) = 0 = f(3). f'(x) = 2x − 4 = 0 ⇒ c = 2, which lies in (1, 3). Hence verified.

Exam tip

For maxima/minima word problems, first write the quantity as a function of one variable using the given condition, then differentiate.

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Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 3

Indefinite Integration

Quick summary

Integration is the reverse process of differentiation. This chapter covers standard integrals, integration by substitution, integration by parts, integration using partial fractions, and special integrals involving a² ± x² forms.

Key points

  • If d/dx F(x) = f(x), then ∫ f(x) dx = F(x) + c, where c is the constant of integration.
  • ∫ [f(x) ± g(x)] dx = ∫ f dx ± ∫ g dx; ∫ k f(x) dx = k ∫ f(x) dx.
  • Substitution: put a part of the integrand equal to t so that its derivative also appears.
  • ∫ f'(x)/f(x) dx = log|f(x)| + c; ∫ eˣ [f(x) + f'(x)] dx = eˣ f(x) + c.
  • Integration by parts: ∫ u v dx = u ∫ v dx − ∫ [u' ∫ v dx] dx. Choose u by the LIATE order (Log, Inverse trig, Algebraic, Trig, Exponential).
  • Partial fractions split a proper rational function into simpler fractions. If the degree of the numerator ≥ degree of the denominator, divide first.
  • Integrals of the form ∫ dx/(a x² + b x + c) are done by completing the square.

Important terms

  • Integrand – the function being integrated.
  • Primitive / antiderivative – a function whose derivative is the integrand.
  • Proper rational function – degree of numerator less than degree of denominator.

Formulas

IntegralResult
∫ xⁿ dxxⁿ⁺¹/(n + 1) + c (n ≠ −1)
∫ 1/x dxlogx+ c
∫ eˣ dxeˣ + c
∫ aˣ dxaˣ/log a + c
∫ sin x dx−cos x + c
∫ cos x dxsin x + c
∫ sec²x dxtan x + c
∫ tan x dxlogsec x+ c
∫ sec x dxlogsec x + tan x+ c
∫ dx/(x² + a²)(1/a) tan⁻¹(x/a) + c
∫ dx/(x² − a²)(1/2a) log(x − a)/(x + a)+ c
∫ dx/(a² − x²)(1/2a) log(a + x)/(a − x)+ c
∫ dx/√(a² − x²)sin⁻¹(x/a) + c
∫ dx/√(x² + a²)logx + √(x² + a²)+ c
∫ √(a² − x²) dx(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + c

5 important questions

Q1. Evaluate ∫ 2x/(1 + x²) dx.

Numerator is the derivative of the denominator. Answer: log(1 + x²) + c.

Q2. Evaluate ∫ x eˣ dx.

By parts with u = x, v = eˣ: x eˣ − ∫ 1 · eˣ dx = x eˣ − eˣ + c = eˣ(x − 1) + c.

Q3. Evaluate ∫ dx/((x + 1)(x + 2)).

1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2). Integral = log|x + 1| − log|x + 2| + c = log|(x + 1)/(x + 2)| + c.

Q4. Evaluate ∫ dx/(x² + 4x + 13).

Complete the square: x² + 4x + 13 = (x + 2)² + 9. Integral = (1/3) tan⁻¹((x + 2)/3) + c.

Q5. Evaluate ∫ eˣ (sin x + cos x) dx.

Here f(x) = sin x and f'(x) = cos x. Using ∫ eˣ[f + f'] dx = eˣ f(x) + c, the answer is eˣ sin x + c.

Exam tip

Learn the standard formula table thoroughly; most questions reduce to one of them after a substitution or completing the square. Never forget "+ c".

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Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 4

Definite Integration

Quick summary

A definite integral gives a number: the net value of an integral between two limits. This chapter covers the definite integral as the limit of a sum, the fundamental theorem of integral calculus, methods of evaluation, and important properties of definite integrals.

Key points

  • Fundamental theorem: if ∫ f(x) dx = F(x), then ∫ from a to b of f(x) dx = F(b) − F(a).
  • The constant of integration is not needed in definite integrals.
  • A definite integral can be seen as the limit of a sum of areas of thin rectangles.
  • When substituting, change the limits too.
  • Properties are very useful to evaluate integrals that are hard to do directly, especially the property f(x) → f(a + b − x).
  • For an even function, the integral from −a to a is twice the integral from 0 to a. For an odd function it is zero.

Important terms

  • Lower limit (a) and upper limit (b) – the end values of x.
  • Even function – f(−x) = f(x), e.g. x², cos x.
  • Odd function – f(−x) = −f(x), e.g. x³, sin x.

Formulas

PropertyStatement (all integrals with respect to x)
P1∫ₐᵇ f(x) dx = ∫ₐᵇ f(t) dt
P2∫ₐᵇ f(x) dx = −∫ᵇₐ f(x) dx
P3integral from a to b = integral from a to c + integral from c to b, where a < c < b
P4∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx
P5∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx
P6∫₀²ᵃ f(x) dx = 2∫₀ᵃ f(x) dx if f(2a − x) = f(x); = 0 if f(2a − x) = −f(x)
P7∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx (f even); = 0 (f odd)

5 important questions

Q1. Evaluate the integral of x² from 0 to 3.

[x³/3] from 0 to 3 = 27/3 − 0 = 9.

Q2. Evaluate the integral of sin x from 0 to π/2.

[−cos x] from 0 to π/2 = −cos(π/2) + cos 0 = 0 + 1 = 1.

Q3. Evaluate I = integral of √(sin x)/(√(sin x) + √(cos x)) from 0 to π/2.

Using P5, replacing x by π/2 − x gives I = integral of √(cos x)/(√(cos x) + √(sin x)) over the same limits. Adding both forms: 2I = integral of 1 dx from 0 to π/2 = π/2. So I = π/4.

Q4. Evaluate the integral of x³ cos x from −π/2 to π/2.

f(x) = x³ cos x. f(−x) = −x³ cos x = −f(x), so f is odd. By P7 the integral is 0.

Q5. Evaluate the integral of 1/(1 + x²) from 0 to 1.

[tan⁻¹x] from 0 to 1 = π/4 − 0 = π/4.

Exam tip

If a definite integral looks impossible directly, try property P4/P5 and add the two forms — this trick appears almost every year.

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Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 5

Application of Definite Integration

Quick summary

Definite integrals can find areas of regions bounded by curves. This chapter covers the area under a curve with the x-axis or y-axis, the area of standard curves (circle, ellipse, parabola), and the area between two curves.

Key points

  • Area bounded by y = f(x), the x-axis and lines x = a, x = b is ∫ from a to b of y dx (when f(x) ≥ 0).
  • Area bounded by x = g(y), the y-axis and lines y = c, y = d is ∫ from c to d of x dy.
  • If the curve is below the x-axis, the integral is negative; take its absolute value for the area.
  • If the curve crosses the axis, split the integral at the crossing points and add absolute values.
  • Area between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to b is ∫ from a to b of [f(x) − g(x)] dx.
  • Find the limits by solving the equations of the curves together (points of intersection).
  • Use symmetry to simplify: for a circle or ellipse, find the area of one quadrant and multiply by 4.
  • Always draw a rough sketch first.

Important terms

  • Bounded region – region enclosed completely by curves and lines.
  • Points of intersection – where the curves meet; they give the limits.
  • Symmetric region – region identical on both sides of an axis.

Formulas

RegionArea
Under y = f(x), a to bA = ∫ₐᵇ y dx
Between curvesA = ∫ₐᵇ [f(x) − g(x)] dx
Circle x² + y² = a²π a²
Ellipse x²/a² + y²/b² = 1π a b
Parabola y² = 4ax and its latus rectum x = a8a²/3
Useful integral∫ √(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)

5 important questions

Q1. Find the area bounded by y = x², the x-axis and the lines x = 1, x = 3.

A = ∫ from 1 to 3 of x² dx = [x³/3] = 27/3 − 1/3 = 26/3 sq units.

Q2. Find the area of the circle x² + y² = a² by integration.

By symmetry, A = 4 ∫ from 0 to a of √(a² − x²) dx = 4[(x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a)] from 0 to a = 4 × (a²/2)(π/2) = π a² sq units.

Q3. Find the area of the ellipse x²/a² + y²/b² = 1.

y = (b/a)√(a² − x²). A = 4(b/a) ∫ from 0 to a of √(a² − x²) dx = 4(b/a)(π a²/4) = π a b sq units.

Q4. Find the area enclosed between the parabola y = x² and the line y = x.

Intersection: x² = x ⇒ x = 0, 1. The line is above the parabola in (0, 1). A = ∫ from 0 to 1 of (x − x²) dx = 1/2 − 1/3 = 1/6 sq units.

Q5. Find the area bounded by y = sin x and the x-axis from x = 0 to x = 2π.

sin x is positive on (0, π) and negative on (π, 2π). A = |∫ from 0 to π of sin x dx| + |∫ from π to 2π of sin x dx| = 2 + 2 = 4 sq units.

Exam tip

A sketch earns marks and avoids errors with signs. Write "sq units" with every answer.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 6

Differential Equations

Quick summary

A differential equation (DE) contains derivatives of an unknown function. This chapter covers order and degree, formation of DEs, solving DEs by variable separation, homogeneous DEs, linear DEs, and applications such as growth and decay, Newton's law of cooling, and surface area/population problems.

Key points

  • Order is the order of the highest derivative in the equation. Degree is the power of the highest order derivative after making the equation free of radicals and fractions in derivatives.
  • A general solution contains arbitrary constants equal in number to the order. A particular solution has specific values of the constants.
  • Formation of DE: differentiate the given relation as many times as there are arbitrary constants, then eliminate them.
  • Variable separable: write f(y) dy = g(x) dx and integrate both sides.
  • Homogeneous DE: dy/dx = f(y/x). Put y = v x, so dy/dx = v + x dv/dx, then separate variables.
  • Linear DE: dy/dx + P y = Q (P, Q functions of x). Integrating factor IF = e^(∫ P dx). Solution: y × IF = ∫ Q × IF dx + c.
  • Growth and decay: dx/dt = k x gives x = x₀ e^(kt) (k > 0 growth, k < 0 decay).
  • Newton's law of cooling: dθ/dt = −k(θ − θ₀).

Important terms

  • Order – highest derivative present.
  • Degree – power of the highest order derivative.
  • Integrating factor (IF) – factor that makes a linear DE exactly integrable.
  • Arbitrary constant – constant c in the general solution.

Formulas

TypeMethod
Variable separable∫ f(y) dy = ∫ g(x) dx + c
Homogeneousy = v x, dy/dx = v + x dv/dx
Linear in yIF = e^(∫ P dx), y·IF = ∫ Q·IF dx + c
Linear in xdx/dy + P x = Q, IF = e^(∫ P dy)
Growth/decayx = x₀ e^(kt)

5 important questions

Q1. Find the order and degree of (d²y/dx²)³ + (dy/dx)⁴ + y = 0.

Highest derivative is d²y/dx², so order = 2. Its power is 3, so degree = 3.

Q2. Form the DE by eliminating the arbitrary constant in y = A e^(2x).

dy/dx = 2A e^(2x) = 2y. DE: dy/dx − 2y = 0.

Q3. Solve dy/dx = (1 + y²)/(1 + x²).

Separate: dy/(1 + y²) = dx/(1 + x²). Integrate: tan⁻¹y = tan⁻¹x + c.

Q4. Solve dy/dx + y/x = x².

P = 1/x, IF = e^(log x) = x. y·x = ∫ x² · x dx = x⁴/4 + c. So x y = x⁴/4 + c.

Q5. A population doubles in 50 years. In how many years will it become triple? (log 2 = 0.6931, log 3 = 1.0986)

dP/dt = kP ⇒ P = P₀ e^(kt). 2 = e^(50k) ⇒ k = log 2/50. For 3P₀: t = log 3/k = 50 × 1.0986/0.6931 ≈ 79.25 years.

Exam tip

Identify the type of DE before solving. For linear DEs, check the coefficient of dy/dx is 1 before finding P and Q.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 7

Probability Distributions

Quick summary

A random variable assigns a number to each outcome of a random experiment. This chapter covers discrete and continuous random variables, probability mass function (p.m.f.), cumulative distribution function (c.d.f.), probability density function (p.d.f.), and the expected value, variance and standard deviation of a random variable.

Key points

  • A random variable X is a real-valued function on the sample space.
  • Discrete random variable: takes countable values (number of heads, number on a die).
  • Continuous random variable: takes any value in an interval (height, time).
  • p.m.f. of a discrete X: P(X = xᵢ) = pᵢ, with each pᵢ ≥ 0 and Σ pᵢ = 1.
  • c.d.f.: F(x) = P(X ≤ x). It is non-decreasing, from 0 to 1.
  • p.d.f. of a continuous X: f(x) ≥ 0 and the integral of f(x) over the whole range = 1. P(a ≤ X ≤ b) = ∫ from a to b of f(x) dx.
  • For a continuous variable, P(X = a) = 0 for any single value a.
  • Expected value (mean) E(X) = Σ xᵢ pᵢ.
  • Variance Var(X) = E(X²) − [E(X)]² = Σ xᵢ² pᵢ − μ². Standard deviation σ = √Var(X).

Important terms

  • Random variable – a numerical outcome of a random experiment.
  • Probability distribution – list of values of X with their probabilities.
  • Expectation – long-run average value of X.
  • Variance – measure of spread of X around its mean.

Formulas

QuantityFormula
Condition on p.m.f.Σ P(X = xᵢ) = 1
c.d.f.F(x) = P(X ≤ x)
MeanE(X) = μ = Σ xᵢ pᵢ
VarianceVar(X) = Σ xᵢ² pᵢ − μ²
SDσ = √Var(X)
p.d.f. conditionintegral of f(x) over its range = 1
Continuous c.d.f.F(x) = integral of f(t) from the lower end to x

5 important questions

Q1. A coin is tossed twice. Let X be the number of heads. Find the probability distribution of X.

S = {HH, HT, TH, TT}. X = 0, 1, 2 with P = 1/4, 2/4, 1/4.

Q2. Find the mean and variance of X in Q1.

E(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1. E(X²) = 0 + 1/2 + 4/4 = 3/2. Var(X) = 3/2 − 1 = 1/2. SD = 1/√2.

Q3. The p.m.f. of X is P(X = x) = k x for x = 1, 2, 3, 4. Find k.

k(1 + 2 + 3 + 4) = 1 ⇒ 10k = 1 ⇒ k = 1/10.

Q4. Verify that f(x) = 2x for 0 < x < 1 (and 0 otherwise) is a p.d.f. Find P(X < 1/2).

f(x) ≥ 0 and ∫ from 0 to 1 of 2x dx = [x²] = 1. So it is a p.d.f. P(X < 1/2) = ∫ from 0 to 1/2 of 2x dx = 1/4.

Q5. Find the expected value of the number shown on a fair die.

E(X) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5.

Exam tip

Always check that probabilities add to 1 before using them. Write the distribution as a neat table with x, p, x p and x² p columns.

📘 Official textbook PDF

Textbook · free PDF, ebalbharati.in

Class 12 (Science, Arts) · Mathematics and Statistics Part 2 · Chapter 8

Binomial Distribution

Quick summary

The binomial distribution gives the probability of a number of successes in a fixed number of independent trials, each with the same probability of success. This chapter covers Bernoulli trials, the binomial distribution, and its mean, variance and standard deviation.

Key points

  • A Bernoulli trial has exactly two outcomes: success (probability p) and failure (probability q = 1 − p).
  • Conditions for binomial distribution: (1) fixed number of trials n; (2) trials are independent; (3) each trial has two outcomes; (4) p is the same in every trial.
  • If X ~ B(n, p), P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ, x = 0, 1, 2, …, n.
  • Mean = n p; variance = n p q; SD = √(n p q).
  • Variance is always less than the mean in a binomial distribution (since q < 1).
  • "At least one" problems: P(X ≥ 1) = 1 − P(X = 0) = 1 − qⁿ.
  • The probabilities are the terms of the binomial expansion of (q + p)ⁿ.

Important terms

  • Success / failure – the two possible outcomes of a trial.
  • Parameters – n and p, which fully describe a binomial distribution.
  • Independent trials – the result of one trial does not affect another.

Formulas

QuantityFormula
p.m.f.P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
MeanE(X) = n p
VarianceVar(X) = n p q
SDσ = √(n p q)
At least oneP(X ≥ 1) = 1 − qⁿ

5 important questions

Q1. A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads.

n = 6, p = q = 1/2. P(X = 4) = ⁶C₄ (1/2)⁶ = 15/64.

Q2. A die is thrown 4 times. Find the probability of getting at least one six.

p = 1/6, q = 5/6. P(X ≥ 1) = 1 − (5/6)⁴ = 1 − 625/1296 = 671/1296.

Q3. For a binomial distribution, mean = 6 and variance = 2. Find n and p.

n p = 6, n p q = 2 ⇒ q = 1/3, p = 2/3. n = 6/(2/3) = 9.

Q4. Find the mean and SD of the number of heads when a coin is tossed 100 times.

n = 100, p = q = 1/2. Mean = 50. Variance = 25. SD = 5.

Q5. Can a binomial distribution have mean 4 and variance 5? Why?

Variance/mean = q = 5/4 > 1, which is impossible since q ≤ 1. So no.

Exam tip

Identify n, p and q clearly first. For "at least" and "at most" questions, use the complement whenever it saves work.

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