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

i think it would be 42cm

i do not know the answer

step-by-step explanation: