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00:00:00 Introduction: Overview of counting, conditional probability, random variables, PMF, PDF, and CDF 00:03:17 Problem 4.10 (Gambler’s Winnings): Using the PMF to find P(X equals 1, 2, or 3 given X is greater than 0) 00:11:21 Counting (Block Permutations): Seating Americans, French, and British so each nationality sits together 00:14:02 Counting (Indistinguishable Objects): Solving the seating problem assuming same nationalities are identical 00:17:03 Inclusion–Exclusion Principle: Probability of buying none or exactly one of suit, shirt, or tie 00:28:26 Joint PDF: Finding the normalizing constant c in f(x,y) equals x divided by 5 plus c times y 00:32:08 Independence Test: Checking whether random variables X and Y are independent 00:38:41 Joint Probability: Calculating P(X plus Y greater than 3) using double integration 00:41:48 Conditional PDF: Finding f(x given y) and computing P(X between 0 and 0.5 given Y equals 2) 00:45:10 Joint Probability Region: Computing P(X between 0 and 1.2 and Y between 0.5 and 3.6) 00:49:48 Problem 4.17 (CDF to PMF): Finding P(X equals 1), P(X equals 2), and P(X equals 3) from a mixed CDF 00:57:17 Problem 4.7 (Dice): Defining random variables for maximum, minimum, sum, and difference of two dice 01:04:48 Problem 4.14 (Poisson): Hurricanes with mean 5.2 per year, probability of three or fewer, and probability of at least one 01:09:04 Binomial–Poisson Application: Defining a safe year and finding probability of two safe years in five 01:15:33 Counting (Multiplication Rule): Selecting a parent and child from different family structures 01:20:39 Problem 3.4 (Bayes’ Theorem): Urn transfer problem and conditional probability 01:27:50 Problem 4.18 (PMF Transformation): Four coin flips, finding PMF of Y equals X minus 2 01:33:50 Problem 3.11 (Hypergeometric): Distribution of defective TVs purchased 01:37:00 Counting (Seating Restrictions): Probability of alternating boys and girls versus girls sitting together 01:41:13 Tree Diagram: Coin flips until HH or TT appears, PMF of number of flips 01:46:06 Problem 5 (Normal Distribution): IQ probabilities for scores above 125 and between 90 and 110