Respuesta :
The probability of person being a heavy smoker given person died is 0.57 .
Baye's Theorem helps in finding the probability of any event A given that event B has already occurred.
For Example the probability of event A given that event B has already occurred is denoted by
[tex]P(A/B)=\frac{P(B/A)P(A)}{P(B)}[/tex]
In the given question
Let H denoted heavy smokers , L denote Light smokers , N denote Non smokers.
Given the probability of non smokers P(N) = 50%=0.5
the probability of light smokers P(L)=20%=0.2
the probability of heavy smokers P(H)=30%=0.3
let P(D/H) denoted the person died given he was a heavy smoker.
let P(D/L) denoted the person died given he was a light smoker.
let P(D/N) denoted the person died given that he was a non smoker.
According to question
"heavy smokers twice as likely to die as light smokers" means
P(D/H)=2P(D/L)......(1)
and "light smokers twice as likely to die as nonsmokers" means
P(D/L)=2P(D/N)
On simplifying we get
[tex]P(D/N)=\frac{1}{2} P(D/L)\\ \\ P(D/N)=0.5 P(D/L)[/tex].....(2)
According to Baye's Theorem
[tex]P(H/D)=\frac{P(D/H)P(H)}{P(D/N)P(N)+P(D/L)P(L)+P(D/H)P(H)}[/tex]
Substituting values of P(D/H) from ...(1) and values of P(D/N) from ....(2)
we get [tex]P(H/D)=\frac{2*P(D/L)P(H)}{0.5*P(D/L)P(N)+P(D/L)P(L)+2*P(D/L)P(H)}[/tex]
taking P(D/L) common from each term
[tex]P(H/D)=\frac{P(D/L)}{P(D/L)} (\frac{2*P(H)}{0.5*P(N)+P(L)+2*P(H)})[/tex]
[tex]P(H/D)= (\frac{2*P(H)}{0.5*P(N)+P(L)+2*P(H)})[/tex]
Substituting the values of P(N), P(H), P(L) from above we get
[tex]P(H/D)= (\frac{2*(0.3)}{0.5*(0.5)+(0.2)+2*(0.3)})[/tex]
[tex]P(H/D)= (\frac{0.6}{0.25+0.2+0.6})[/tex]
=[tex]\frac{0.6}{1.05}=0.57[/tex]
Therefore , the probability of person being a heavy smoker given person died is 0.57 .
Learn more about Probability here https://brainly.com/question/17764757
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