数学代写|STAT230 Probability Theory

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STAT230课程简介

This course provides an introduction to probability models including sample spaces, mutually exclusive and independent events, conditional probability and Bayes’ Theorem. The named distributions (Discrete Uniform, Hypergeometric, Binomial, Negative Binomial, Geometric, Poisson, Continuous Uniform, Exponential, Normal (Gaussian), and Multinomial) are used to model real phenomena. Discrete and continuous univariate random variables and their distributions are discussed. Joint probability functions, marginal probability functions, and conditional probability functions of two or more discrete random variables and functions of random variables are also discussed. Students learn how to calculate and interpret means, variances and covariances particularly for the named distributions. The Central Limit Theorem is used to approximate probabilities.Note: Many upper-year Statistics courses require a grade of at least 60% in STAT 230. Offered: F,W,S

Prerequisites 

This course uses the inverted classroom approach. This means that students will either watch video lectures and/or read book chapters before class. Class will be used for exercises and examples that highlight the core concepts in the material. To ensure that students go through materials before class there is quiz to take online that needs to be completed the day before each lecture. The class exercises will be group exercises to help solidify concepts in the lectures and give some examples of how these concepts are used. The exercises will be done in small groups (3-4) and will typically involve pencil-and-paper work. Sometimes you will also run some code (in Rstudio) which will be provided in advance to help illustrate ideas.

STAT230 Probability Theory HELP（EXAM HELP， ONLINE TUTOR）

Sure, the sample space for this experiment, where a fair coin is tossed three times, would be:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Where H represents heads and T represents tails.

The probability of getting two heads can be found by adding the probabilities of all outcomes that have exactly two heads. From the sample space {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}, we can see that there are three outcomes with exactly two heads: HHT, HTH, and THH.

The probability of getting two heads is therefore:

P(Two heads) = P(HHT) + P(HTH) + P(THH)

Since each coin toss is independent and has a probability of 1/2 of resulting in heads, the probability of each outcome is:

P(HHT) = (1/2) * (1/2) * (1/2) = 1/8

P(HTH) = (1/2) * (1/2) * (1/2) = 1/8

P(THH) = (1/2) * (1/2) * (1/2) = 1/8

Therefore, the probability of getting two heads is:

P(Two heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8

So the probability of getting two heads is 3/8 or 0.375.

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