In a mail system 20% of all messages are spam. 70% of spam messages contain the word "lottery", while only 12.5% of legitimate messages contain it. A message arrives containing the word "lottery". What is the probability that it is spam?

Step-by-step solution

1. 1

   Name the events and sort the given numbers. Let $G$ = "the message is spam", $N$ = "the message is legitimate", and $L$ = "the message contains the word lottery". The data are

   $P(G)=0.2,\qquad P(N)=0.8$
   $P(L\mid G)=0.7,\qquad P(L\mid N)=0.125$

   Notice the direction: you are given the probability of the word given the class, but asked for the class given the word. Reversing a conditional probability is exactly what Bayes' theorem is for.

2. 2

   Write down Bayes' theorem for the quantity asked.

   $P(G\mid L)=\frac{P(L\mid G)\,P(G)}{P(L)}$

   The numerator is the probability of being spam and containing the word; the denominator is the probability of containing the word at all, over the whole mail stream.

3. 3

   Get the denominator from the law of total probability. Every message is either spam or legitimate, so those two cases partition the sample space:

   $P(L)=P(L\mid G)P(G)+P(L\mid N)P(N)=0.7\times 0.2+0.125\times 0.8$
   $P(L)=0.14+0.10=0.24$

   So about one message in four contains the word — and most of that mass comes from the far larger pool of legitimate mail.

4. 4

   Substitute and simplify to an exact fraction.

   $P(G\mid L)=\frac{0.14}{0.24}=\frac{14}{24}=\frac{7}{12}\approx 0.5833$

5. 5

   Interpret the result and sanity-check it. Seeing the word raises the probability of spam from the prior $20\%$ to about $58.3\%$ — strong evidence, but still far from certainty, because legitimate mail is four times as common to begin with. A useful cross-check: of a notional $1000$ messages, $200$ are spam and $140$ of those contain the word, while $800$ are legitimate and $100$ of those contain it. Of the $240$ flagged messages, $140$ are spam: $140/240=7/12$, matching the formula.