According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that men who were very active (burning about 3,500 calories daily) were a fifth as likely to develop diabetes compared with men who were sedentary. Assume that one-fourth of all middle-aged men are very active, and the rest are classified as sedentary. What is the probability that a middle-aged man with diabetes is very active? (Round your answer to four decimal places.)
$$
P(a|d)=\frac{P(d|a)p(a)}{P(d)}
$$
$P(d)=\frac{202}{5990}$
Could someone break down for me how to do this one? I have done all the other ones on my homework but this one has me stuck as I am not sure what to categorize as what
Best Answer
You want $P(a|d)$ where $a$ means active, $d$ means diabetic. Then Baye's rule gives $$ P(a|d)=\frac{P(d|a)p(a)}{P(d)} $$ We know $P(d)=\frac{202}{5990}$. $P(a)=1/4$ and $P(d|a)$ we need to figure out. But $$ P(d)=1/4P(d|a)+3/4p(d|s)=1/4P(d|a)+3/4*5P(d|a) $$ Can you finish from here?