Bayes' rule can be used to identify and filter spam emails and text messages.In this collection, 747 of the 5574 total messages (13.40%) are identified as spam. The word “free” is contained in 4.75% of all messages, and 3.57% of all messages both contain the word “free” and are marked as spam. The word “text” (or “txt”) is contained in 7.01% of all messages, and in 38.55% of all spam messages. Of all spam messages, 17.00% contain both the word “free” and the word “text” (or “txt”). Of all non-spam messages, 0.06% contain both the word “free” and the word “text” (or “txt”) . Given that a message contains the word “free” but does NOT contain the word “text” (or “txt”), what is the probability that it is spam?

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Asked Sep 13, 2019
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Bayes' rule can be used to identify and filter spam emails and text messages.In this collection, 747 of the 5574 total messages (13.40%) are identified as spam. The word “free” is contained in 4.75% of all messages, and 3.57% of all messages both contain the word “free” and are marked as spam. The word “text” (or “txt”) is contained in 7.01% of all messages, and in 38.55% of all spam messages. Of all spam messages, 17.00% contain both the word “free” and the word “text” (or “txt”). Of all non-spam messages, 0.06% contain both the word “free” and the word “text” (or “txt”) . Given that a message contains the word “free” but does NOT contain the word “text” (or “txt”), what is the probability that it is spam?

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Expert Answer

Step 1

Given Data

P(Spam) = 0.1340

P(Not spam) = 1- P(spam) = 1 - 0.1340 = 0.866

P(Free) = 0.0475

P(Free and spam) = 0.0357

P(Text) = 0.0701

P(Text and Free if Spam) = 0.17

Step 2

Solving

By Applying conditional probability

Р(Free n spaт)
P(Spam)
Free
0.0357
0.2664
\Spam/
Using The rule of total probability
Р(Free nText)
0.134
(Text n Free
- P(Spam) х Р
Spam
Text n Free
+ P(пot Spam)ХР
Spam
0.1340 x 0.17 0.866 X 0.0006
0.0233
P(Free n not Text) P(Free) - P(Free U Text)
0.0475 0.0233 = 0.0242
Free n Text
- P
Free n not Text
P
Free
0.2664
0.1700 = 0.0964
\Spam/
Spam
Spam
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Image Transcriptionclose

Р(Free n spaт) P(Spam) Free 0.0357 0.2664 \Spam/ Using The rule of total probability Р(Free nText) 0.134 (Text n Free - P(Spam) х Р Spam Text n Free + P(пot Spam)ХР Spam 0.1340 x 0.17 0.866 X 0.0006 0.0233 P(Free n not Text) P(Free) - P(Free U Text) 0.0475 0.0233 = 0.0242 Free n Text - P Free n not Text P Free 0.2664 0.1700 = 0.0964 \Spam/ Spam Spam

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Step 3

Now Applying Baye...

Free n not Text
P(Spam) х Р
Spam
Spam
P
Free n not Text/
Р(Free n not Теxt)
0.134 x 0.0964
0.534
0.0242
help_outline

Image Transcriptionclose

Free n not Text P(Spam) х Р Spam Spam P Free n not Text/ Р(Free n not Теxt) 0.134 x 0.0964 0.534 0.0242

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