Given : In a head injury,
Probability of having Skull Fracture
$P(F)=6\%=0.06$
$\therefore$ Probability of not having Skull Fracture
$P(\bar F)=1-P(F)=0.94$
Probability of having Nausea when person doesn’t have Skull fracture
$P(N|\bar F)=70\%=0.70$
$\therefore$ Probability of not having Nausea when person doesn’t have Skull fracture
$P(\bar N|\bar F)=1-P(N|\bar F)=0.30$
Probability of having Nausea when person have Skull Fracture,
$P(N|F)=98\%=0.98$
$\therefore$ Probability of not having Nausea when person have Skull fracture
$P(\bar N|F)=1-P(N| F)=0.02$
Question : Probability of having Skull Fracture when person is not Nauseous, $P(F|\bar{N})$
By Bayes theorem ,
$P(F|\bar N)=\frac {P(\bar N|F)\times P(F)}{P(\bar N)}$
$P(\bar N)=P(\bar N|F)\times P(F)+P(\bar N|\bar F)\times P(\bar F)$
$=0.02\times0.06+0.3\times0.94=0.2832$
$\therefore P(F|\bar N)=\frac {0.02\times 0.06}{0.2832}=\frac {0.0012}{0.2832}=\frac1 {236}\approx 0.0042$
PS :
This question reminded me of something similar,
https://www.youtube.com/watch?v=lG4VkPoG3ko
refer this in case Bayes’ theorem interests you :)
