WILL WARD

Warning: Spoilers Follow From Stat 110, Chapter 1 Ex 1.31 Elk dwell in acertain forest. There are N elk, of which a simple random sample of size n are captured and tagged ("simple random sample" means that all sets of n elk are equally likely). The captured elk are returned to the population, and then a new sample is drawn, this time with size m. This is an important method that is widely used in…

Warning: Spoilers Follow Ex 2.14: Consider the following scenario, from Tversky and Kahneman: Let A be the event that before the end of next year, Peter will have installed a burglar alarm system in his home. Let B denote the event that Peter's home will be burglarized before the end of next year. (a) Intuitively, which do you think is bigger, or ? Explain your intuition. We're not given any…

Warning: Spoilers Follow From Unit 2 of Blitzstein/Hwang Probability (Harvard's Stat 110 probability course) Ex. 2.11 An exit poll in an election is a survey taken of voters just after they have voted. One major use of exit polls has been so that news organizations can try to figure out as soon as possible who won the election, before the votes are officially counted. This has been notoriously…

Note: This puzzle dates back to the 1950s, so old-fashioned assumptions were made. This is stated in the textbook as well. (a) A couple has two children. The older child is a girl. What is the probability that both children are girls? Not knowing anything else, we there are four ways in which the couple could have had two children (Here, we use the convention that the letter on the left represents…

Warning: Spoilers Follow 1. How many ways are there to permute the letters in the word MISSISSIPPI? If each letter were to have its own unique ID and it were the case that only the IDs mattered, there would be 11! permutations of the word MISSISSIPPI. However, were not permuting unique IDs, but letters, some of which get used several times, which means that 11! permutations would include…

Warning: Spoilers Follow A homework problem from Harvard's Stat 110 course has two chess players ready to play 7 games. The possible outcomes for each game are Win (1 point), Draw (0.5 points), and Loss (0 points). If all seven games were to be played regardless of the outcomes of the individual games, it's possible that one player could have 7 points and the other 0 points. We'll call the two…

Calculating the probability of a full house is shown in chapter 1 of Harvard's Stat 110 (Probability) course, and is pretty straightforward. First, get the number of ways to choose three of a kind, multiply it by the number of ways to choose a pair from the remaining cards, and divide that product by the total number of possible hands. The calculation is shown below. which gives ~0.00144. A small…

Here's an experimental verification of the theoretical probability given in the last post: Poker Hand Probabilities: One Pair