Let's continue with the previous scenario: 0.03 0.97 Y N Y Dc 0.98 N D 0.99 0.01 0.02 i) The probability of having the disease and testing positive is P(DNY)=P(D) P(Y|D) = Number ii) The probability of not having the disease and testing positive is P(DCnY) = P(DC) P(Y|DC) = Number iii) The total probability of testing positive is, by the Total Probability Rule, the sum of these, namely P(Y) = P(D) P(Y|D) + P(DC) P(Y|DC) = Number
Let's continue with the previous scenario: 0.03 0.97 Y N Y Dc 0.98 N D 0.99 0.01 0.02 i) The probability of having the disease and testing positive is P(DNY)=P(D) P(Y|D) = Number ii) The probability of not having the disease and testing positive is P(DCnY) = P(DC) P(Y|DC) = Number iii) The total probability of testing positive is, by the Total Probability Rule, the sum of these, namely P(Y) = P(D) P(Y|D) + P(DC) P(Y|DC) = Number
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Let's continue with the previous scenario:
0.03
0.97
D 0.99. Y
0.01 N
0.02
Y
DC 0.98 N
i) The probability of having the disease and testing positive is
P(DnY)=P(D) P(Y|D) = Number
ii) The probability of not having the disease and testing positive is
P(DC Y) = P(DC) P(Y|DC) = Number
iii) The total probability of testing positive is, by the Total Probability Rule, the
sum of these, namely
P(Y) = P(D) P(Y|D) + P(DC) P(Y|DC) = Number
The conditional probability of D given Y is, by Bayes' Rule,
P(Dn Y)
P(Y)
P(DY) =
Number
So there is roughly a 40%
chance that you do not have the
disease even if you do test positive for it. This is an important and perhaps
surprising result that is worth understanding: just because you test positive for
something does not guarantee that you actually have it.
Note: your answers should be entered with at least 2 significant figures of
accuracy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feba998f7-aba2-4966-983b-2f872110f909%2Ff66f56de-3b57-4273-983f-6b7cfe61f1b4%2Fgcjse5v_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let's continue with the previous scenario:
0.03
0.97
D 0.99. Y
0.01 N
0.02
Y
DC 0.98 N
i) The probability of having the disease and testing positive is
P(DnY)=P(D) P(Y|D) = Number
ii) The probability of not having the disease and testing positive is
P(DC Y) = P(DC) P(Y|DC) = Number
iii) The total probability of testing positive is, by the Total Probability Rule, the
sum of these, namely
P(Y) = P(D) P(Y|D) + P(DC) P(Y|DC) = Number
The conditional probability of D given Y is, by Bayes' Rule,
P(Dn Y)
P(Y)
P(DY) =
Number
So there is roughly a 40%
chance that you do not have the
disease even if you do test positive for it. This is an important and perhaps
surprising result that is worth understanding: just because you test positive for
something does not guarantee that you actually have it.
Note: your answers should be entered with at least 2 significant figures of
accuracy.
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