Question

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of the valid messages. Also, 20% of the messages are spam.
1. Determine the following probabilities:
a. The message contains free.
b. The message is spam given that it contains free.
c. The message is valid given that it does not contain free.

Answer 1

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

• The probability of the word free occurring in a spam message is,             $P(F|S)=0.60$
• The probability of the word free occurring in a valid message is,             $P(F|V)=0.04$
• The probability of spam messages is,

        $P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

Answer 2

The answer are 15.2%, 12% and 96.8% respectively.

Given to us:

word free occurs in the spam messages is 60%,

word free occurs in the valid messages is 4%,

20% of the messages are spam,

to simplify the question, let us assume that a total of x number of messages are there.

So,

20% of the messages are spam = $0.20 \times x$

80% of the messages are valid = $0.80 \times x$

a.) Probability of the message contains free;

Since, 60% of the spam messages contain word Free

Probability of the spam message contains free

= percentage of the messages that are spam x percentage of the messages         that are spam containing word free

$=( 0.20)\times (0.60)\times x\\= 0.12 \times x$

Also, 4% of the valid messages contain word Free;

Probability of the valid message contains free

= percentage of the messages that are valid x percentage of the messages that are valid containing word free

$=( 0.80)\times (0.04)\times x\\= 0.032\times x$

Probability of the message contains free

= Probability of the spam message contains free + Probability of the valid message contains free

=0.12x + 0.032x

= 0.152x

= 15.2% of  x

Therefore, the probability of the message contains free is 15.2%.

b.) Probability of the message is spam given that it contains free;

Since, 60% of the spam messages contain word Free

Probability of the spam message contains free

= percentage of the messages that are spam x percentage of the messages that are spam containing word free

$=( 0.20)\times (0.60)\times x\\= 0.12 \times x$

Therefore, the probability of the message contains free is 12%.

c.) Probability of the message is valid given that it does not contain free,

Since, 4% of the valid messages contain word Free;

Probability of the valid message contains free

= percentage of the messages that are valid x percentage of the messages         that are valid containing word free

$=( 0.80)\times (0.04)\times x\\= 0.032\times x$

Also, we know that the probability of any sure event is 100% = 1.

So, Probability of the message is valid which does not contain free

= 1 - Probability of the valid message contains free

= 1 - 0.032

= 0.968

= 96.8% of x.

Therefore, Probability of the valid message that does not contain free is

96.8%.

Hence, the answer are 15.2%, 12% and 96.8% respectively.

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