Answer
$29 \%$
Thus, we conclude the probability that youths arrested or charged with a crime would have no preschool education is approximately $29 \%$.This means that youths who had preschool education would be less to be arrested or charged with a crime than who would have no preschool education.
Work Step by Step
According to Bayes' theorem:
$P(N|C)=\dfrac{P(C|N)P(N)}{P(C|N)P(N)+P(C|N')P(N')}~~~~~~~~(1)$
Here, we have
$ P(C|N)=0.51\\ P(C'|N')=0.31\\ P(N')=0.8$
Now, we will now use formula (1) and the given data to obtain:
$P(N|C)=\dfrac{P(C|N)P(N)}{P(C|N)P(N)+P(C|N')P(N')}=\dfrac{(0.51)(1-0.8)}{(0.51)(1-0.8)+(0.31)(0.8)}$
or, $ \approx 0.29$
or, $=29 \%$
Thus, we conclude the probability that youths arrested or charged with a crime would have no preschool education is approximately $29 \%$.This means that youths who had preschool education would be less to be arrested or charged with a crime than who would have no preschool education.
