In a large collection of real SMS text messages from particular cellphone users, 13.40% of messages are identified as spam. Of all spam messages 17.0% contain both the word "free" and the word "text" ("txt") For example, "Congrats!! You have been selected to receive a free camera phone, txt ******* to claim your prize." Of all non=spam messages, 6% contain both the words "free" and the word "text" (or "txt"). Given that a message contains both the word "free" and the word "text" (or "txt"), what is the probability that it is spam?
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To find the probability that a message is spam given that it contains both the words "free" and "text" (or "txt"), we can use Bayes' theorem.
Let's define the following events:
A: Message is spam
B: Message contains both the words "free" and "text" (or "txt")
We are given the following information:
P(A) = 0.134 (probability of a message being spam)
P(B|A) = 0.17 (probability of a message containing both the words "free" and "text" given that it is spam)
P(B|A') = 0.06 (probability of a message containing both the words "free" and "text" given that it is not spam)
We need to find P(A|B) (probability of a message being spam given that it contains both the words "free" and "text").
Using Bayes' theorem, the equation is as follows:
P(A|B) = (P(B|A) * P(A)) / P(B)
To calculate P(B), we need to consider both spam and non-spam messages:
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
First, we calculate P(A') using the complement rule:
P(A') = 1 - P(A)
= 1 - 0.134
= 0.866
Now we can calculate P(B):
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
= 0.17 * 0.134 + 0.06 * 0.866
Finally, we can calculate P(A|B) using the formula mentioned earlier:
P(A|B) = (P(B|A) * P(A)) / P(B)
= (0.17 * 0.134) / (0.17 * 0.134 + 0.06 * 0.866)
Evaluating the last expression gives us the probability that a message is spam given that it contains both the words "free" and "text" (or "txt").