Little exercise on conditional probability

Question

I would like to know if I solved this exercise correct:

2/3 of your e-mails ends up in the spam folder, while the rest goes into your normal mail inbox. 99% of the mails in the spam folder are spam and 1% is not spam.

In your mail inbox you have 20% spam mails and 80% normal mails.

1. What is the total amount of spam you receive?

2. What is the probability that a normal mail ends up in the spam folder?

1. $$(2/3\cdot 99)/100 +(1/3\cdot 20)/100 = 1/15 + 33/50 = 109/150$$

2. Let $A$ be the event 'in spam folder' and $B$ be the event 'normal mail'

Then we have $$P(B | A) = P(A \cap B)/P(A) = (2/3)\cdot(1/100)\cdot (3/2)=1/100$$

Answer 1

$\frac 23 (\frac {99}{100}) + \frac 13 (\frac {20}{100}) = \frac {218}{300}$ of your total mail is spam.

The easy way to do part b) is to assume you receive exactly 300 messages. How many are normal? 100 arrive in your inbox. 80 of which are normal. 200 go to spam, 2 of which are normal.

$\frac 2{82} $ of your normal e-mail ends up in the spam folder.

If you want to use Bayes law.

$A$ is the probability a message is in your spam folder.

$B$ is the probability the message is normal.

$P(B) = 1 - \frac {218}{300} = \frac {82}{300}$

$P(B|A) = P(A|B) \frac {P(A)}{P(B)}\\ P(B|A) = \frac {1}{100} \frac{\frac {2}{3}}{\frac{82}{300}} =\frac {2}{82}$