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 June 25th, 2018, 09:35 AM #1 Newbie   Joined: Jun 2018 From: Toronto Posts: 2 Thanks: 0 Probability Question Please solve this basic question, at-least 1 part- Suppose we have a data set with two variables: type of injury (categorical) and description (string). We want to predict the type of injury for new data given only the description. We get a new description that contains the word “swelling”. Our model, built from a very large training sample, tells us that the only two types of injuries that can produce the word “swelling” are “Burn”, which occurs in 1 out of 10 observations, and “Bruise”, which occurs in 1 out of 100 observations. A “Bruise” observation has a 30% chance of generating the word “swelling”, while “Burn” has only a 5% chance of generating the word “swelling". 1. Without any other information, is the new observation with the word “swelling” more likely to be a burn or a bruise? What is the probability of either? * 2. What is the probability of at-least 2 bruises given that the 6 observations have descriptions that contain the word “swelling”. I solved the answer as - GIVEN- Burn - 1 out of 10 observations Bruise - 1 out of 100 observations Bruise - 30% chance of generating word swelling Burn - 5% change of generating word swelling CALCULATION 1 - So, 0.01*.3 = .003 = P(bruise) .1*.05 = .005 = P(burn) 6 observations have the word "swelling". CALCULATION 2 - P(at least 2 bruises) = P(2 bruises, 4 burns) + P(3 bruises, 3 burns) + P(4 bruises, 2 burns) + P(5 bruises, 1 burns) + P (6 bruises, 0 burns) = 9*625+27*125+81*25+243*5+729 / 1000*1000 Is it correct? Please help. Thanks & Best
 June 26th, 2018, 10:34 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,099 Thanks: 1093 let $B$ stand for Bruise $N$ stand for Burn $S$ stand for Swelling we are given $P[B]=0.1$ $P[N]=0.01$ $P[S|B]=0.3$ $P[S|N]=0.05$ We want to solve for and compare $P[B|S]$ and $P[N|S]$ This is just an application of Bayes Rule. $P[B|S] = \dfrac{P[S|B]P[B]}{P[S]}$ $P[N|S] = \dfrac{P[S|N]P[N]}{P[S]}$ where $P[S] = P[S|B]P[B] + P[S|N]P[N]$ dumping your numbers into this we get $P[S] = 0.0305$ $P[B|S] = 0.983607$ $P[N|S] = 0.0163934$ and thus a Bruise is much more likely given the description Swelling
 June 26th, 2018, 10:44 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,099 Thanks: 1093 for the second calculation they ask for $P[B\geq 2|S=6]$ $P[B\geq 2|S=6] = \dfrac{P[S=6|B\geq 2]P[B\geq 2]}{P[S=6]}$ $P[B\geq 2]$ is easy enough to compute using the binomial distribution $P[S=6]$ is a bit trickier but straightforward enough. $P[S=6|B\geq 2]$ is just a subset of the terms used to find $P[S=6]$ see if you can work through it.

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