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hw6_solutions

# hw6_solutions - CSE 630 Homework 6 Solutions Key 1(12...

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CSE 630: Homework 6 Solutions Key 1. (12 points) Consider a game where two six-sided dice are being rolled. (a) What is the total number of atomic events in this game? Answer: 36. (b) What is the probability of getting a 4 on one of the two dice? Answer: 11/36 (not 12/36). There are 11 atomic events with a 4 on one of the two dice, amongst a total of 36. By the principle of indifference, it is reasonable to assume that each atomic event is equally likely, and hence the final probability is 11/36. (If you thought the answer was 12/36, you are double counting the event with 4s on both dice.) (c) Now, suppose I tell you that the sum of the two dice is greater than 7. What is the probability of getting a 4 on one of the two dice now? Answer: One can calculate this directly using the definition of conditional proba- bility. The number of atomic events where the sum of the two dice is greater than 7 is 15 (by explicit counting). The number of events within this set of 15 events with a 4 is 5 (again by explicit counting). So the joint probability of getting a 4 on one of the dice as well as sum over 7 is 5/36, and the probability of the sum of the two dice being greater than 7 is 15/36; the conditional probability of getting a 4 on one of the dice given that the sum is greater than 7 is 5/15 = 1/3. One can also calculate this using Bayes’ Rule. We already saw in (b) that the prior probability of obtaining a 4 on one of the dice is 11/36; now we want the posterior probability given that the sum is greater than 7. To apply Bayes’ Rule we will need the data (or evidence) likelihood, which is the probability of sum greater than 7, given that one of the dice is 4. There are 5 events with sum greater than 7 among the 11 events which have at least one 4 on them; hence the evidence likelihood is 5/11. The prior probability of the evidence is the probability of sum being greater than 7, which is 15/36. By Bayes’ rule, posterior probability is prior probability times evidence likelihood divided by prior probability of evidence. Hence, posterior probability of 4 on one of the two dice given sum is greater than 7 is 11 / 36 * 5 / 11 15 / 36 = 5 / 15 = 1 / 3. 2. (16 points) After your yearly checkup, the doctor has bad news and good news. The bad news is that you tested positive for a serious disease and that the test is 99% accurate (i.e., the probability of testing positive when you do have the disease is 0.99, as is the probability of testing negative when you don’t have the disease). The good news is that this is a rare disease, striking only 1 in 10,000 people of your age. What are the chances that you actually have the disease? Answer: First thing we should do is to define our random variables. Let T be the boolean random variable which is true if the test is positive and false otherwise. Let D be the boolean random variable which is true if you have the disease and false otherwise.

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