Dr. Brajesh Kumar Jha

Discrete and Continuous Probability Functions 1. Discrete Probability Function Let \(X\) be a discrete random variable. A function \[ p(x)=P(X=x) \] is called the probability mass function (PMF) or discrete probability function of \(X\). For a function \(p(x)\) to be a valid discrete probability function, it must satisfy the following conditions: \[ \boxed{p(x)\geq 0} \] and \[ \boxed{\sum_x p(x)=1}. \] Thus, the probability of an event such as \(a\leq X\leq b\) is obtained by adding the corresponding probabilities: \[ P(a\leq X\leq b) = \sum_{x=a}^{b}p(x). \] 5 Solved Examples on Discrete Probability Functions Example 1: Determine the Constant in a PMF A discrete random variable \(X\) has probability function \[ P(X=x)=kx,\qquad x=1,2,3,4. \] Find: The value of \(k\) \(P(X\leq 2)\) The mean \(E(X)\) Show Solution Since the total probability must be equal to 1, \[ \sum_{x=1}^{4} P(X=x)=1. \] Therefore, \[ k...

Random Variable: Definition, Types and 10 Solved Examples Definition of Random Variable A random variable is a real-valued function defined on the sample space of a random experiment. \( X:S\rightarrow \mathbb{R} \) It assigns a numerical value to every possible outcome of a random experiment. Example Suppose two coins are tossed. The sample space is \( S=\{HH,HT,TH,TT\}. \) Let \(X\) denote the number of heads obtained. Then \(X(HH)=2,\quad X(HT)=1,\quad X(TH)=1,\quad X(TT)=0.\) Therefore, \(X\in\{0,1,2\}.\) Types of Random Variables 1. Discrete Random Variable A random variable is called discrete if it takes a finite or countably infinite number of values. Its probability mass function (PMF) is \(p_X(x)=P(X=x)\) and it satisfies \(p_X(x)\geq 0,\qquad \sum_x p_X(x)=1.\) 2. Continuous Random ...

Functions and Inverse Functions – 10 Practice Questions Question 1. Determine whether the relation R = {(1,2), (2,4), (3,6), (4,8)} defines a function. Show Solution A relation is a function if every input has exactly one output. Here the inputs are 1, 2, 3, and 4, and each input has only one corresponding output. Therefore, R is a function. Question 2. Given f(x) = 3x - 5 find f(2), f(-1), and f(0). Show Solution For x = 2: f(2) = 3(2) - 5 = 6 - 5 = 1 For x = -1: f(-1) = 3(-1) - 5 = -3 - 5 = -8 For x = 0: f(0) = 3(0) - 5 = -5 Hence, f(2) = 1, f(-1) = -8, and f(0) = -5. Question 3. If f(x) = x 2 + 2x - 3 find f(a + 1). Show Solution Replace x by a + 1: f(a + 1) = (a + 1) 2 + 2(a + 1) - 3 Expanding, = a 2 + 2a + 1 + 2a + 2 - 3 = a 2 + 4a Therefore, f(a + 1) = a 2 + 4a. Question 4. Find the domain of f(x) = (2x + 1)/(x - 4) Show Solution The denominator cannot be zero. x - 4 ≠ 0 Therefore, x ≠ 4 Hence...

📘 Mathematics-I Interactive notes on mathematical functions, their domains, ranges, properties and animated graphs. Functions and Their Graphs A function describes a relationship between an input x and an output f(x) . Different functions produce characteristic curves. Click any function below to expand the section. The graph will then be drawn automatically. 1. Exponential Functions eˣ Exponential Function ⌄ f(x) = e x Domain ℝ Co-domain ℝ Range (0, ∞) The value of e x is always positive. The x-axis is a horizontal asymptote. ▶ Replay Graph 2ˣ Exponential Function ⌄ f(x) = 2 x Domain ℝ Co-domain ℝ Range (0, ∞) The graph passes through (0,1) and increases rapidly as x increases. ▶ Replay Graph 2. Trigonometric Functions sin Sine Function ⌄ f(x) = sin x Domain ℝ Co-domain ℝ Range ...

Bayes' Theorem Definition, Formula and 10 University-Level Solved Problems Definition of Bayes' Theorem Bayes' theorem is used to calculate the probability of an event after new information or evidence has become available. Let A and B be two events such that P(B) > 0. Then Bayes' theorem is P(A | B) = [P(B | A) P(A)] / P(B) Here: P(A | B) = posterior probability P(B | A) = likelihood P(A) = prior probability P(B) = probability of the evidence Prior Likelihood Posterior P(A) P(B | A) P(A | B) Figure 1: Bayes' theorem updates prior probability using new evidence. General Form If A 1 , A 2 , ..., A n form a partition of the sample space, then P(A i | B) = [P(A i ) P(B | A i )] / Σ P(A j ) P(B | A j ) Example 1: Two Machines A factory has two machines M 1 and M 2 . Machine M 1 produces 60% of the items and M 2 produces 40%. Their defective rates are 2% and 5%, respect...