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Assignment: Probability and Statistics Basic

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Sticky Ad Probability Problems with Detailed Solutions Click each question to expand the detailed interpretation and solution. Question 1: Two unbiased dice are thrown Two fair dice are rolled together. Find the probability that: (i) the sum is even (ii) the sum is less than 5 (iii) the sum is at most 12 Since each die has 6 possible outcomes, the total number of outcomes is: $$6\times6=36$$ We calculate the number of favorable outcomes for each condition. (i) Sum is even The sum is even when: both numbers are even both numbers are odd Number of even numbers on a die = 3 Number of odd numbers on a die = 3 $$3\times3+3\times3=18$$ Therefore, $$P(\text{sum is even})=\frac{18}{36}=\frac12$$ (ii) Sum is less than 5 Possible sums are 2, 3 and 4. Number of favorable outcomes: $$1+2+3=6$$ H...
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Some Questions on Differential Equations

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Differential Equations: Detailed Step-by-Step Solutions In this article, we solve three important differential equations step by step using standard methods such as complementary function and particular integral , integrating factor , and separation of variables . 1) Solve \(y''+4y=\sin(3x)\) This is a linear differential equation with constant coefficients: \[ y''+4y=\sin(3x) \] The general solution is: \[ y=y_c+y_p \] where \(y_c\) is the complementary function and \(y_p\) is the particular integral. Step 1: Complementary Function The auxiliary equation is: \[ m^2+4=0 \] \[ m^2=-4 \] \[ m=\pm 2i \] Therefore, the complementary function is: \[ y_c=C_1\cos 2x+C_2\sin 2x \] Step 2: Particular Integral \[ (D^2+4)y=\sin 3x \] \[ y_p=\frac{1}{D^2+4}\sin 3x \] Using: \[ f(D)\sin ax=f(-a^2)\sin ax \] \[ y_p=\frac{1}{-9+4}\sin 3x \] \[ y_p=-\frac15\sin 3x \] Step 3: General Solution \[ y=C_1\cos 2x+C_2\sin 2x-\frac15\sin 3x \] Fi...
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Chapter 3: First Order Homogeneous Differential Equations

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First Order Homogeneous Differential Equations (Solved Examples) A differential equation of the form \[ \frac{dy}{dx}=f\left(\frac{y}{x}\right) \] is called a homogeneous differential equation . We use the substitution \[ y=vx \quad \text{or} \quad v=\frac{y}{x} \] so that \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Example 1: Solve \( \frac{dy}{dx}=\frac{x+y}{x} \) View Solution \[ \frac{dy}{dx}=1+\frac{y}{x} \] Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=1+v \] \[ x\frac{dv}{dx}=1 \] \[ \frac{dv}{dx}=\frac{1}{x} \] Integrate \[ v=\ln|x|+C \] Since \(v=\frac{y}{x}\) \[ \frac{y}{x}=\ln|x|+C \] \[ y=x(\ln|x|+C) \] Example 2: Solve \( \frac{dy}{dx}=\frac{x+y}{x-y} \) View Solution Let \[ v=\frac{y}{x}, \quad y=vx \] \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute \[ v+x\frac{dv}{dx}=\frac{1+v}{1-v} \] \[ x\frac{dv}{dx}=\frac{1+v}{1-v}-v \] \[ x\frac{dv}{dx}=\frac{1+v^2}{...
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ODE Chapter 1: First Order Differential Equations using Separation of Variables

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First Order Differential Equations using Separation of Variables Below are solved examples of first order differential equations using the separation of variables method . Example 1: Solve \( \frac{dy}{dx}=3x^2 \) View Solution Separate variables \[ dy = 3x^2 dx \] Integrate both sides \[ \int dy = \int 3x^2 dx \] \[ y = x^3 + C \] Example 2: Solve \( \frac{dy}{dx}=xy \) View Solution Separate variables \[ \frac{dy}{y}=x\,dx \] Integrate \[ \int \frac{1}{y}dy = \int x\,dx \] \[ \ln |y| = \frac{x^2}{2}+C \] \[ y = Ce^{x^2/2} \] Example 3: Solve \( \frac{dy}{dx}=\frac{x}{y} \) View Solution Separate variables \[ y\,dy = x\,dx \] Integrate \[ \int y\,dy = \int x\,dx \] \[ \frac{y^2}{2} = \frac{x^2}{2} + C \] \[ y^2 = x^2 + C \] Example 4: Solve \( \frac{dy}{dx}=y^2 \) View Solution Separate variables \[ \frac{dy}{y^2}=dx \] Integrate \[ \int y^{-2}dy = \int dx \] \[ -\frac{1}{y}=x+C \] \[ y=\frac{1}{C-x} ...
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Differential Equation: Application

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First Order Differential Equations using Separation of Variables Advanced Application Problems (Separation of Variables) The following examples illustrate applications of first order differential equations in population growth, cooling law, chemical reactions, and physics. Example 11 (Population Growth): If population grows according to \( \frac{dP}{dt}=kP \), find \(P(t)\). View Solution Separate variables \[ \frac{dP}{P}=k\,dt \] Integrate \[ \int \frac{dP}{P}=\int k\,dt \] \[ \ln P = kt + C \] \[ P = Ce^{kt} \] Example 12 (Newton's Law of Cooling): \[ \frac{dT}{dt}=-k(T-T_s) \] Find \(T(t)\). View Solution Separate variables \[ \frac{dT}{T-T_s}=-k\,dt \] Integrate \[ \ln |T-T_s|=-kt+C \] \[ T-T_s=Ce^{-kt} \] \[ T=T_s+Ce^{-kt} \] Example 13 (Radioactive Decay): Solve \( \frac{dN}{dt}=-kN \) View Solution \[ \frac{dN}{N}=-k\,dt \] Integrate \[ \ln N=-kt+C \] \[ N=Ce^{-kt} \] Example 14 (Chemical Reaction): So...
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Cholesky Decomposition Method

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Cholesky Decomposition Method – Solved Problems The Cholesky Decomposition Method is used to solve systems of linear equations when the coefficient matrix is symmetric and positive definite . The matrix is decomposed as A = LL T where L is a lower triangular matrix. The solution is obtained in two steps: Forward substitution: LY = B Backward substitution: L T X = Y Problem 1 Solve using Cholesky decomposition 4x + 2y = 6 2x + 3y = 7 Show Solution Matrix form A = [4 2 2 3] L = [2 0 1 √2] Forward substitution: 2y₁ = 6 y₁ = 3 3 + √2 y₂ = 7 y₂ = 2√2 Backward substitution: √2 y = 2√2 y = 2 2x + 2 = 3 x = 1/2 Solution: x = 1/2 y = 2 Problem 2 Solve 4x + 2y + 2z = 8 2x + 5y + z = 3 2x + y + 3z = 5 Show Solution Matrix A = [4 2 2 2 5 1 2 1 3] Cholesky factor L = [2 0 0 1 2 0 1 0 √2] Forward substitution gives Y = [4 -1/2 1/√2] Backwar...
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Total Derivatives

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Total Derivatives – Theory and Solved Problems Definition If \(z=f(x,y)\) where \(x\) and \(y\) depend on parameter \(t\), then the total derivative is \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] This formula is known as the **Chain Rule for multivariable functions**. Visual Understanding of the Chain Rule In problems involving total derivatives , variables often depend on other variables. The Chain Rule shows how a change in one variable affects another through intermediate variables. The following diagrams illustrate this dependency structure. Diagram 1: Basic Chain Rule Structure z x y t This diagram represents: \[ z = f(x,y), \quad x = x(t), \quad y = y(t) \] \[ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \] Diagram 2: Two-Level Dependency z x y u...
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