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Among individuals who have sustained head injuries, x-rays reveal that only about 6% have skull fractures. Nausea is a standard symptom of a skull fracture and occurs in 98% of all skull fracture cases with other types of head injuries, nausea is present in about 70% of all cases. Suppose that an individual who has just suffered a head injury is not nauseous. Find the probability that he has a skull fracture _____

Corrected Question statement :

Among individuals who have sustained head injuries, x-rays reveal that only about 6% have skull fractures. Nausea is a standard symptom of a skull fracture and occurs in 98% of all skull fracture cases . With other types of head injuries, nausea is present in about 70% of all cases. Suppose that an individual who has just suffered a head injury is not nauseous. Find the probability that he has a skull fracture _____

I am just adding the Tree method here to solve this type of question. Hope it helps.

You just need to follow the terms mentioned in the question and will get the tree for sure. The answer mentioned below is correct as well.

.ll.

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,

refer this in case Bayes’ theorem interests you :)