Your e-mail provider has a very good spam filter that only 1.5% of all e-mails p

Question

Your e-mail provider has a very good spam filter that only 1.5% of all e-mails passed the filter are spam. A lot of spam e-mails are trying to phish for your personal information so they ask you to go to an unknown web site where you are asked to enter this information. About 98% of the spam email will ask you to click on a link without providing you with the explicit url. In addition, 3% of the legitimate e-mails will ask you to click on a link without the explicit url being shown. For a random e-mail in your inbox (so this email has passed the filter), if you are provided a link without the explicit url, what is the probability that the e-mail is spam? Why is your answer counterintuitive?

Explanation / Answer

Solution

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..………………….(3)

Now, to work out the solution,

Let A represent the event that the email is spam and B represent the event that the email asks to go to a link without specifying the url.

Given that only 1.5% of all emails passing through the filter are spam, if the mail is in inbox,

=> P(A) = 0.015 …………………………………………………………………………..(4)

(4) => P(AC) = P(email is not a spam) = 0.985………………………………………….(5)

Given that 98% of spam emails ask for going to a link of unknown url

=> P(B/A) = 0.98 ………………………………………………………………(6)

Further given that 3% of legitimate emails ask for going to a link of unknown url

=> P(B/AC) = 0.03 ………………………………………………………………(7)

Now, we want to find the probability that the email is spam given that the email asked for going to a link of unknown url. i.e., we want to find P(A/B).

Vide (3) under Back-up Theory,

P(A/B) = P(B/A) x { P(A)/P(B)}

= 0.98 x {0.015/P(B)} [substituting from (6) and (4)]…………………………………(8)

Now, Vide (2) under Back-up Theory,

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}

= (0.98 x 0.015) + (0.03 x 0.985) [substituting from (6), (4), (7) and (5)]

= 0.0147 + 0.02955

= 0.4425 ……………………………………………………………………………(9)

(8) and (9) => P(A/B) = {(0.98 x 0.015)/0.4425}

= 0.0332 ANSWER

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