What’s the addition expression that has the value of 8

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

ANEK [815]

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.

4 0

3 years ago

How did the graph of f(x) = x^2 and g(x) = 3/4 x^2 relate?

nirvana33 [79]

g(x) = $\frac{3}{4} f(x)$ or g(x) is 3/4 times of f(x) , F(x) and g(x) have common solution or intersecting point in the graph parabola at x=0 i.e. in origin and x = $\frac {4}{3}$ .

Step-by-step explanation:

We have a function f(x) = $x^{2}$ and another function , g(x) = $\frac{3}{4} x^{2}$ . In the graph of y = $x^{2}$ , the point (0, 0) is called the vertex. The vertex is the minimum point in a parabola that opens upward. In a parabola that opens downward, the vertex is the maximum point.

Graphing y = (x - h)2 + k , where h = 0 & k = 0

Function g(x) can be formed with compression in function f(x) by a factor of 3/4 , i.e. g(x) = $\frac{3}{4} f(x)$ or g(x) is 3/4 times of f(x).Domain and range of f(x) and g(x) are same ! Although structure of both functions is same the only difference is g(x) is compressed vertically by a factor 3/4. Both are graph of a parabola with vertex at (0,0). Also, F(x) and g(x) have common solution or intersecting point at x=0 i.e. in origin.