s) An e-mail lter is planned to separate valid e-mails from spam. The word free occurs in 50% of the spam messages and only 3% of the valid messages. Also, 20% of the messages are spam. Determine the following probabilities: (a) The message contains the word free. (b) The message is spam given that it contains free. (c) The message is valid given that it does not contain free.

A) P (F) = 0.124

B) P (S|F) = 0.8065

C) P (V|F^ (c)) = 0.886

Step-by-step explanation:

Let us denote as follows;

F = Message contains word free

S = message is spam

V = message is valid

From the question, we are given that;

The probability that word free occurs in spam messages; P (F|S) = 50% = 0.5

The probability of the valid messages that contain free; P (F|V) = 3% = 0.03

Spam messages; P (S) = 20% = 0.2

Valid messages; P (V) = 1 - 0.2 = 0.8

A) From rule of total probability;

probability that the message contains the word free is given as;

P (F) = P (F|S) •P (S) + P (F|V) •P (V)

P (F) = (0.5 x 0.2) + (0.03 x 0.8)

P (F) = 0.124

B) From Baye's theorem;

probability that the message is spam given that it contains free is given as;

P (S|F) = P (F|S) •P (S) / P (F)

P (S|F) = (0.5 x 0.2) / 0.124

P (S|F) = 0.8065

C) From combination of complement rule and Baye's theorem;

probability that the message is valid given that it does not contain free is given as;

P (V|F^ (c)) = P (F^ (c) |V) •P (V) / P (F^ (c))

Thus,

P (V|F^ (c)) = [ (1 - P (F|V)) •P (V) ] / (1 - P (F))

P (V|F^ (c)) = ((1 - 0.03) •0.8) / (1 - 0.124)

P (V|F^ (c)) = 0.776/0.876

P (V|F^ (c)) = 0.886