Bayes Theorem - Having trouble applying it I'm trying to apply Bayes Theorem in a assignment's problem but I'm having trouble with it.

Here is the question:

"A insurance company classify its insureds into two risk categories: 80% of them are classified as "low risk" and 20% of them are classified as "high risk". The probability that one "low risk" insured dies is 0.1 and the probability that one "high risk" insured dies is 0.5. We know that one person died this year. What is the probability that this person was classified as "high risk"?" The answer is $\frac{5}{9}$

Here is what I did:

First, we have two events:

A: Being a person classified as high risk and 
B: Die

So, the problem asks P(A|B), correct?

I did like that, so: P(A|B) = $\frac{P(A\bigcap B)}{P(B)}$ = $\frac{P(B|A)P(A)}{P(B)}$ = $\frac{0.5 * 0.2}{0.1*0.8 + 0.5*0.2} = \frac{0.1}{0.18}$

P(B|A) = 0.5 (it was given in the exercise), P(A) = 0.2 (it was given too) and P(B) = 0.1 * 0.8 + 0.5 * 0.2 = 0.18

What I did wrong? The correct answer is $\frac{5}{9}$
P(H|D) = probability that person is high risk given that he is described as "high risk"

By bayes theorem

P(H|D) = (P(H)*(D|H))/P(D)
P(H) = 0.20
P(D|H) = 0.5

P(D) = P(H)P(D|H) + P(L)P(D|L) = 0.20*0.5 + 0.80*0.1

0.20*0.5/(0.20*0.5 + 0.80*0.1) = 5/9
P(H|D) = probability that person is high risk given that he is described as "high risk"

By bayes theorem

P(H|D) = (P(H)*(D|H))/P(D)
P(H) = 0.20
P(D|H) = 0.5

P(D) = P(H)P(D|H) + P(L)P(D|L) = 0.20*0.5 + 0.80*0.1

0.20*0.5/(0.20*0.5 + 0.80*0.1) = 5/9