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Chapter 11: Problem 62

Of all spam messages, \(17.00 \%\) contain both the word "free" and the word "text" (or "txt"). For example, "Congrats!! You are selected to receive a free camera phone, \(t x t\) * to claim your prize." Of all non-spam messages, \(0.06 \%\) contain both the word "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?

Short Answer

Expert verified
The probability that a message is spam, given it contains the words 'free' and 'text (or txt)', is approximately \(99.65%\) (or 0.9965 in decimal form).

Step by step solution

01

Defining the Events

Let's label the events. Let's say, A is the event 'message is spam' and B is the event 'message contains the words free and text (or txt).' We are given \(P(A) = 0.17\) and \(P(\overline{A}) = 0.0006\), where \(\overline{A}\) is the event 'message is not spam'.
02

Calculate the Total Probability of Event B

The total probability of a message containing the words 'free' and 'text (or txt)' regardless if it's spam or not is given by the total probability rule: \(P(B)= P(A)P(B|A) + P(\overline{A})P(B|\overline{A}) = (0.17) \cdot (1) + (0.0006) \cdot (1) = 0.1706\). Considering the assumption that if either 'spam' or 'non-spam' contains the words 'free' and 'text (or txt)', that's one certain event.
03

Applying Bayes Theorem

We need to compute \(P(A|B)\), which is the probability that a message is spam given that it contains both 'free' and 'text (or txt)'. For that, we can use Bayes theorem: \(P(A|B) = \dfrac{P(A)P(B|A)}{P(B)} = \dfrac{0.17 \cdot 1}{0.1706} = 0.9964705882352941\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a fundamental concept in mathematics that quantifies uncertainty. It is a measure of the likelihood of a specific event occurring out of the total possible outcomes. In the context of our exercise, we are focusing on the probability of spam messages containing certain words.

To break it down:
  • The probability of an event is a value between 0 and 1, where 1 signifies certainty, and 0 signifies impossibility.
  • The probability that a message is spam is denoted as \(P(A)\), while the probability that a message is not spam is \(P(\overline{A})\).
  • In our exercise, we have \(P(A) = 0.17\) representing spam likelihood.
Understanding probability helps us make predictions and informed decisions based on available data.
Spam Detection
Spam detection is an essential process in email filtering systems, used to identify and block unwanted messages. One common technique to detect spam is to analyze the keywords in messages. In our example, messages containing the words "free" and "text" (or "txt") are examined.

Here's how it works in practice:
  • Certain combinations of words, such as "free" and "text", are more common in spam messages, as seen in the given data where 17% of spam messages contain these words.
  • Only a tiny fraction (0.06%) of non-spam messages contain these words.
Using these patterns allows email systems to apply probabilistic models to improve spam filtering accuracy and enhance user experience.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. This is an essential element of Bayes' Theorem, used in our exercise.

Let's consider:
  • In the exercise, the conditional probability \(P(A|B)\) is described as the likelihood that a message is spam, given it contains the keywords "free" and "text".
  • This is found using Bayes' Theorem: \(P(A|B) = \frac{P(A)P(B|A)}{P(B)}\).
  • Here, \(P(B)\) is the overall probability of a message containing these words, calculated using the total probability rule.
Conditional probability provides a more refined analysis by incorporating new conditions or findings into our predictions.

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Most popular questions from this chapter

We have a bag of peanut \(M \&\) M's with \(80 \mathrm{M} \&\) Ms in it, and there are 11 red ones, 12 orange ones, 20 blue ones, 11 green ones, 18 yellow ones, and 8 brown ones. They are mixed up so that each is equally likely to be selected if we pick one. (a) If we select one at random, what is the probability that it is yellow? (b) If we select one at random, what is the probability that it is not brown? (c) If we select one at random, what is the probability that it is blue or green? (d) If we select one at random, then put it back, mix them up well (so the selections are independent) and select another one, what is the probability that both the first and second ones are red? (e) If we select one, keep it, and then select a second one, what is the probability that the first one is yellow and the second one is blue?

Use the information that, for events \(\mathrm{A}\) and \(\mathrm{B},\) we have \(P(A)=0.8, P(B)=0.4\) and \(P(A\) and \(B)=0.25\). Are events \(A\) and \(B\) independent?

Use the probability function given in the table to calculate: (a) The mean of the random variable (b) The standard deviation of the random variable $$ \begin{array}{llll} \hline x & 10 & 20 & 30 \\ \hline p(x) & 0.7 & 0.2 & 0.1 \\ \hline \end{array} $$

An Internet Service Provider (ISP) offers its customers three options: \- Basic: Standard internet for everyday needs, at \(\$ 23.95\) per month. \- Premium: Fast internet speeds for streaming video and downloading music, at \(\$ 29.95\) per month. Ultra: Super-fast internet speeds for online gaming, at \(\$ 39.95\) per month. Ultra is the company's least popular option; they have twice as many Premium customers, and three times as many Basic customers: (a) Let \(X\) be the monthly fee paid by a randomly selected customer. Give the probability distribution of \(X\). (b) What is the mean of \(X ?\) (This is the company's average monthly revenue per customer.) (c) What is the standard deviation of \(X ?\)

Calculate the requested binomial probability. Find \(P(X=3\) ) if \(X\) is a binomial random variable with \(n=10\) and \(p=0.4\)
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