Please answer part 2 Part 2: A website reports that 70% of its users are from outside a certain country. Out of their usersfrom outside the country, 60% of them log on every day. Out of their users from inside the country,80% of them log on every day.(a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a).(b) Using Bayes' Theorem, out of users who log on every day, what is the probability that theyare from inside the country? Part 1: Suppose that F and X are events from a common sample space with P(F) #0 and P(X) + 0.(a) Prove that P(X) = P(X|F)P(F) + P(X|F)P(F). Hint: Explain why P(X|F)P(F)P(X nF) is another way of writing the definition of conditional probability, and then usethat with the logic from the proof of Theorem 4.1.1.

Answered: Part 2: A website reports that 70% of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Please answer part 2

Part 2: A website reports that 70% of its users are from outside a certain country. Out of their users
from outside the country, 60% of them log on every day. Out of their users from inside the country,
80% of them log on every day.
(a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a).
(b) Using Bayes' Theorem, out of users who log on every day, what is the probability that they
are from inside the country?

Part 1: Suppose that F and X are events from a common sample space with P(F) #0 and P(X) + 0.
(a) Prove that P(X) = P(X|F)P(F) + P(X|F)P(F). Hint: Explain why P(X|F)P(F)
P(X nF) is another way of writing the definition of conditional probability, and then use
that with the logic from the proof of Theorem 4.1.1.

Expert Solution

Step 1 PART 1

Let S denote the sample space of events X and F.

then note that $S = F U F$ and $X = X \cap S$

Then ,

$X = X \cap (F \cup \bar{F}) X = (X \cap F) \cup (X \cap \bar{F}) . . . . . . (1)$

Result : If A and B are events of a sample space S ,then

$P ((A U B)) = P (A) + P (B) - P ((A \cap B))$

Let $A = X \cap F$ and $B = X \cap \bar{F}$ then $X = A U B$ ( by using (1))

And $P (X) = P ((A U B))$

note that $A \cap B = (X \cap F) \cap (X \cap \bar{F}) = X \cap (F \cap \bar{F}) = \emptyset$ (as $F \cap \bar{F}$ is an empty set )

which implies $P (A \cap B) = 0$

therefore ,

$\begin{array}{rcl} P (X) & = & P ((A U B)) \\ P (X) & = & P (A) + P (B) \\ P (X) & = & P (X \cap F) + P (X \cap \bar{F}) . . . . . . . . . (2) \end{array}$

By definition of conditional probability , $P (X | F) = \frac{P (X \cap F)}{P (F)}, P (F) \neq 0$

So we have , $P (X \cap F) = P (F) P (X | F) . . . . . (3)$ ,provided $P (F) \neq 0$

Also, $P (X \cap \bar{F}) = P (\bar{F}) P (X | \bar{F}) . . . . . . . . . (4)$

( note that $P (\bar{F}) \neq 0$ as $P (\bar{F}) = 1 - P (F)$ and $0 < P (F) < 1$ )

Now ,using (2), (3) and (4) , we get

$\begin{array}{rcl} P (X) \end{array} = P (F) P (X | F) + P (\bar{F}) P (X | \bar{F})$

Hence proved .

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