DAT 520 Problem Set 3.docx - DAT 520 Problem Set 3...

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DAT 520 Problem Set 3 Cumulative and Conditional Probability Overview: The last problem set in Module Two exposed you to dependent trials that caused the probability to shift slightly with each choice of sock from the drawer. Kind of sneaky, eh? This problem set gets you a little further into conditional probability with a direct look at Bayes’ theorem. Since this is a course in decision analysis and not a course in probability, we are going to introduce these concepts, use them once or twice by hand, but then let R and Excel do the heavy lifting for us the rest of the time. This problem set, however, requires hand calculations. All you should need is a calculator and your wits. Example Problem: Example 1 : You are flipping a fair coin four times in total, but you are flipping it in groups of two. In order to flip the coin for flips three and four, you have to get heads two times in a row for flips one and two. Problem Notes: Define success: Success for this problem is flipping a head, which is p =50%. Recall the binomial equation: Recall the Law of Total Probability equation: probability(item_1) * (item_1_%contribution) + probability(item_2) * (item_2_ %contribution) + … Recall Bayes’ theorem: Questions for Example Problem: Q1: What is the probability of flipping heads four times in a row, generally ? .
Q2: The way this problem is set up, what is the total probability of getting four heads? :
We calculate this as 25% chance of flipping two heads on flips one and two. Same thing for flips three and four. Let’s now call the first event p (F1&2) and the second one p (F3&4). Next, build the table . Notice that all the % contributions have to add up to 1. Table of contributions to total probability: P % contrib. p (F1&2) .25 .50 p (F3&4) .25 .50 Using the law of total probability : Total probability for two heads then two heads is (.50 x .25) + (.50 x .25) = 25%.

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