tag:blogger.com,1999:blog-6894866515532737257.post3907083582945814725..comments2018-02-19T11:09:02.513-08:00Comments on Probably Overthinking It: My favorite Bayes's Theorem problemsAllen Downeyhttps://plus.google.com/111942648516576371054noreply@blogger.comBlogger19125tag:blogger.com,1999:blog-6894866515532737257.post-23560589514582152542017-02-16T07:39:09.813-08:002017-02-16T07:39:09.813-08:00Excellent question! I just turned it into a blog p...Excellent question! I just turned it into a blog post. I'll give readers a few days before I post a solution: http://allendowney.blogspot.com/2017/02/a-nice-bayes-theorem-problem-medical.htmlAllen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-46894475359124415992017-02-16T04:53:17.056-08:002017-02-16T04:53:17.056-08:00I have a question. Exactly 1/5th of the people in ...I have a question. Exactly 1/5th of the people in a town have Beaver Fever . There are two tests for Beaver Fever, TEST1 and TEST2. When a person goes to a doctor to test for Beaver Fever, with probability 2/3 the doctor conducts TEST1 on him and with probability 1/3 the doctor conducts TEST2 on him. When TEST1 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 3/4. If the person does not have the disease, the result is positive with probability 1/4. When TEST2 is done on a person, the outcome is as follows: If the person has the disease, the result is positive with probability 1. If the person does not have the disease, the result is positive with probability 1/2. A person is picked uniformly at random from the town and is sent to a doctor to test for Beaver Fever. The result comes out positive. What is the probability that the person has the disease?Riyahttps://www.blogger.com/profile/05095295993427070799noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-31672788098146932342016-11-30T13:28:12.526-08:002016-11-30T13:28:12.526-08:00Yes, this "natural frequency" way of sol...Yes, this "natural frequency" way of solving problems like this is excellent.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-43918002372392650512016-11-30T13:16:22.615-08:002016-11-30T13:16:22.615-08:00Simplest way to answer this question is to assume ...Simplest way to answer this question is to assume a number for total population.<br />Assume that 400 is the combined population of all three states. Then each of state1,state2,state3 will be with population 160,100,140 respectively. From the data provided we can derive that number of people supporting party1 in state1,state2,state3 are 0.5*160,0.6*10,0.35*140 (= 80,60,49). <br />From the above derived information we can evaluate the probability that a party1 supporter will be from state2 is 60/(80+60+49) = 0.317 Approx.Unknownhttps://www.blogger.com/profile/04089618677994568066noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-59345034728425643072016-03-04T07:39:16.466-08:002016-03-04T07:39:16.466-08:00Fun problem, thanks! I posted a solution here: ht...Fun problem, thanks! I posted a solution here: https://github.com/AllenDowney/ThinkBayes2/blob/master/code/examples/voter.ipynbAllen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-82938046279938308392016-03-02T22:09:38.339-08:002016-03-02T22:09:38.339-08:00The following data is about a poll that occurred i...The following data is about a poll that occurred in 3 states. In state1, 50% of voters support Party1, in state2, 60% of the voters support Party1, and in state3, 35% of the voters support Party1. Of the total population of the three states, 40% live in state1, 25% live in state2, and 35% live in state3. Given that a voter supports Party1, what is the probability that he lives in state2?amithttps://www.blogger.com/profile/15478781866754575977noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-64052545609886952062015-03-05T11:19:51.181-08:002015-03-05T11:19:51.181-08:00I don't understand the question. It doesn'...I don't understand the question. It doesn't look like you are using Bayes's theorem. Your first and third expressions are equivalent. The middle one is not (unless P(S|~D) = 1, which would be weird).Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-70697695866227202112015-03-05T05:56:59.667-08:002015-03-05T05:56:59.667-08:00Bayes Theorm : In bayes theorm if I know the value...Bayes Theorm : In bayes theorm if I know the value of p(symptoms/disease) let say 0.3 so, is it justifiable if I take p(~symptoms/disease)=(1-p(symptoms/disease)) p(symptoms/~disease) = (1-p(symptoms/disease))?AKANSHA PANDEYhttps://www.blogger.com/profile/00763971001212693605noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-32729520169278376572014-05-28T10:03:55.804-07:002014-05-28T10:03:55.804-07:00That's funny. Thanks!That's funny. Thanks!Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-30015951511152233472014-05-28T09:42:04.241-07:002014-05-28T09:42:04.241-07:00I realize this is an older post, but it is great s...I realize this is an older post, but it is great stuff; I was watching your video on YouTube that covers this material and stopped to work it out myself. On the m&m problem, there are actually two issues: one which you've outlined above, but the second could be stated as, "What are the chances that the dispenser of m&m's is not a good friend, given the early date of those bags of candy?!" http://www.eatbydate.com/other/sweets/mms/<br /><br />Thanks again, your work is much appreciated.<br /><br />SteveSThttps://www.blogger.com/profile/06615807234500868464noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-88304803737373483492013-12-09T15:42:32.284-08:002013-12-09T15:42:32.284-08:00Just fixed it. Thanks!Just fixed it. Thanks!Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-83869492385930501142013-12-09T14:32:18.264-08:002013-12-09T14:32:18.264-08:00Problem 2 should have 24% blue instead of 20%. Th...Problem 2 should have 24% blue instead of 20%. The probabilities did not add up to 1 so I googled the problem to find out which one was off.Chris Kaylinhttps://www.blogger.com/profile/12789472904090726331noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-77637666164227874362013-11-10T07:04:44.138-08:002013-11-10T07:04:44.138-08:00If you Google this problem, you'll find lots o...If you Google this problem, you'll find lots of online solutions. If you are taking a class and don't know how to solve this, ask questions until you do!Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-89737365005812498612013-11-10T05:54:15.977-08:002013-11-10T05:54:15.977-08:00A biometric security device using fingerprints err...A biometric security device using fingerprints erroneously refuses to admit 1 in 1,000 authorized persons from a facility containing classified information. The device will erroneously admit 1 in 1,000,000 unauthorized persons. Assume that 95 percent of those who seek access are authorized. If the alarm goes off and a person is refused admission, what is the probability that the person was really authorized? <br /><br />Please help me solve this problem as soon as possible .plss.....sonu bhttps://www.blogger.com/profile/12735871771743590772noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-78455452455033247112013-10-20T10:59:48.770-07:002013-10-20T10:59:48.770-07:00I agree that we need more information, but I am pr...I agree that we need more information, but I am pretty sure that answer from About.com is wrong. Among fraternal twins, 25% are GG, 25% are BB, and 50% are BG or GB.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-29278387581767702262013-10-20T09:24:15.453-07:002013-10-20T09:24:15.453-07:00Nice post! To answer the Elvis question, you actu...Nice post! To answer the Elvis question, you actually still need some additional information (from About.com): Among fraternal twins, 1/3 are two girls, 1/3 are two boys, and 1/3 are a boy and a girl. Also (just to be perfectly thorough) identical twins are certain to have the same gender.Wild Orchids for Trotskyhttps://www.blogger.com/profile/00844340000411944242noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-37154137771018727732013-10-19T12:29:16.204-07:002013-10-19T12:29:16.204-07:00hi nikhil,compute the probability of drawing 2 pla...hi nikhil,compute the probability of drawing 2 plain cookies from each bowl ie 30C2/40C2 and 20C2/40C2 for 1 and 2 respectively and then use this probabilities to solve using baye's theorem. Kunal Dhttps://www.blogger.com/profile/17037199112105651786noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-53933208515997768872013-04-08T13:17:09.443-07:002013-04-08T13:17:09.443-07:00Hi Nikhil, Good question. Maybe I will post an u...Hi Nikhil, Good question. Maybe I will post an update with a longer answer, but here's the short version. You can extend the analysis shown above to handle this data by computing the likelihood of the data under each hypothesis. For example, the likelihood of drawing two vanilla cookies (without replacement) from a bowl with 20 vanilla and 20 chocolate cookies is (20)(19) / (40)(39), which is a little less than 1/4.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-26113553477410153982013-04-06T09:43:38.142-07:002013-04-06T09:43:38.142-07:00Suppose there are two full bowls of cookies. Bowl ...Suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks TWO cookies at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. BOTH the cookies turns out to be a plain one. How probable is it that Fred picked them out of Bowl #1?<br />I just wanted to know how to solve the problem of the cookies if in the above cookies problem, if two cookies were selected at a time , then what is the probability that both the cookies are from bowl#1??<br />how to solve this, please tell me.nikhilhttps://www.blogger.com/profile/02149675857196298438noreply@blogger.com