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June 25th, 2018, 10:35 AM   #1
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Post 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
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June 26th, 2018, 11:34 AM   #2
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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
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June 26th, 2018, 11:44 AM   #3
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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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