The spam filter and the word "free"

Question

In a certain inbox, 40% of incoming emails are spam and 60% are legitimate. The word "free" shows up in 60% of spam emails but in only 4% of legitimate ones. A new email arrives containing the word "free."

What is the probability that this email is spam?

Accepted Answer

Let S be "spam" and F be "contains free." We want P(S \mid F). Imagine 1000 emails. - Spam: 400 emails; 60\% contain "free" \Rightarrow 240 flagged. - Legitimate: 600 emails; 4\% contain "free" \Rightarrow 24 flagged. Of the 240 + 24 = 264 emails containing "free," the spam ones number 240: P(S \mid F) = \frac{240}{264} = \frac{10}{11} \approx 0.909. By Bayes' theorem this is P(S\mid F) = \frac{0.6 	imes 0.4}{0.6	imes 0.4 + 0.04 	imes 0.6} = \frac{0.24}{0.264} \approx 0.909. Takeaway. Natural frequencies ("out of 1000 emails…") make Bayes' theorem feel obvious. A single strong keyword can …

The spam filter and the word "free"

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