Exercise 2: A Second Spam Filter Example — Possible Solution
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GIVEN
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P(spam)                  = 0.30
P("winner" | spam)       = 0.25
P("winner" | not spam)   = 0.01

STEP 1: THE LAW OF TOTAL PROBABILITY — P("winner")
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P("winner") = P("winner"|spam) x P(spam) + P("winner"|not spam) x P(not spam)
            = (0.25 x 0.30) + (0.01 x 0.70)
            = 0.075 + 0.007
            = 0.082

STEP 2: APPLYING BAYES' THEOREM
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P(spam|"winner") = [P("winner"|spam) x P(spam)] / P("winner")
                  = 0.075 / 0.082
                  ~= 0.9146 (about 91.46%)

INTERPRETATION
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Seeing the word "winner" raises the belief an email is spam from a
30% prior all the way to a roughly 91.5% posterior - a much sharper
update than Chapter 4's own "free" example (which only reached 80%),
since "winner" is a stronger, more spam-specific signal here (25% vs.
1% is a bigger gap than 30% vs. 5%).

WHY THIS WORKS AS AN ANSWER
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The calculation follows this chapter's own two-step Bayes' Theorem
method exactly - first computing the total probability of the
evidence via the law of total probability, then applying Bayes'
Theorem to get the posterior - and the result is interpreted by
comparing it directly against the chapter's own "free" example to
explain why this particular word produces a stronger update.