March, my birthday, and I'm still here!

Hi Readers! I am still here responding to people's requests for homework help, I just don't post the answers on my blog as much anymore. But rest assured, I am still here answering your emails.

Also, to celebrate my birthday this week, I invite you to partake in this fun online gaming experience called "Auditorium"!

http://www.playauditorium.com/

I enjoyed it so much that I played it all the way through on my first sitting!

For M. Weaver's class

Hi guys, here are the solutions to the problems from "Solving Linear Systems by Matrices", starting with #5. Use them to check your work. 

5. x = 6, y = 12

6. x = 3, y = -5

7. x=5, y=-2

8. x=8, y=10

9. x=7, y=10

10. x=7, y=-21

11. x=8, y=6

12. x=9, y=7

13. x=-21, y=3

14. x=-2, y=-5

15. x=3, y=-2

16. x=7, y=11

17. x=8, y=9

18. x=5, y=-21

19. x=-5, y=5

20. x=-5, y=6

21. x=9, y=6

22. x=9, y=-21

23. x=5, y=4

24. x=-21, y=-5

I will post the rest soon (no later than 6 PM).

Updates!

Hi! Sorry to those whose emails I haven't answered recently- I've been receiving a lot of spam disguised as help requests! Anyway, I'll be answering the recent posts on the forums- hehheh, pretty excited to have the first forum posts.

What exactly is a function, anyway?

I recently received an email and one of the problems that the sender was seeking help with was:

"Question 1: Determine which set of ordered pairs (x,y) represents y as a function of x.

a) {(4, -2), (-8, -1), (-8, 4), (-1, -8)}
b) {(4, -2), (-2, -8), (4, -1)}
c) {(4, -2), (-2, 4), (-1, -1)
d) {4, -2, -8, -1}"

I had this pretty cool graphical way of explaining how a function differs from a set of ordered pairs (click on the images to enlarge).

A Diophantine Equation

Since I haven't received any direct requests for homework help lately, I went searching for problems instead. Here is a recent problem I answered for someone on Yahoo! Answers, under my alter-ego (eason13@yahoo.com):

"Find all possible solutions to the following equation:

x+1/(y+1/z)=13/9,

where x, y, z must be natural numbers."

Here is my solution:
"By natural number I assume we mean positive integers.
Well, 1<= 13/9 <2.>= 1 since it is a natural number. Thus, x = 1.

Thus, 1 + 1/(y+1/z) = 13/9
so 1/(y+1/z) = 4/9
so y+1/z = 9/4.
2<= 9/4 < 3, so y is either 1 or 2.

IF y =1:
Then 1/z = 5/4, an impossibility. Thus, y is not 1.

Thus, y = 2.
Then 1/z = 1/4 so z=4.

Thus, the unique solution among natural numbers (positive integers) is x = 1, y = 2, z = 4."

Jeweler's Problem (probability)

Recently I received the following e-mail:

Hi, I wasnt sure how to do this problem, and wanted to know if you can help:

Carison Jewelers permits the return of their diamond wedding rings, provided the return occurs within 2 weeks of the purchase date. Their records reveal that 10% of the diamond wedding rings are returned. If 5 different customers buy 5 diamond rings, what is the probability that none of the rings will be returned? What is the probability that more than 3 rings will be returned?

So do I do .

Probability that none will be returned = Binomialcdf(5,10,5)

Probability that more than 3 will be returned = Binormalcdf(5,10,1)

??? I am confused

thanks for the help

First, let's talk a little about probability. In probability, we have what are called "events". An event has several possible "outcomes". For instance, if your even is a single toin coss, then the possible outcomes are heads and tails.

Probability is the likelihood of a desired outcome occurring. For instance, if you flip a fair coin, the probability of getting heads is 1/2.

We write the probability of achieving an outcome "x" from an event "E" as:

P(E=x)

Now, there are some more laws of probability we need to learn. First of all, Say we are dealing with two events, one is called "E" and the other is called "F". Say that x is an outcome for E and y is an outcome for F. Then the probability that E=x and F=y (i.e. the probability that we get BOTH outcomes that we want)
can be written

P((E=x) AND (F=y))

Furthermore, it satisfies this relation:

P((E=x) AND (F=y)) = P(E=x)*P(F=y)

(where the asterisk denotes multiplication).
So, for example, if we have two separate coins, then the probability that we get heads on the first AND heads on the second is 1/2*1/2 = 1/4.

Another law that probability satisfies is that

P((E=x) OR (E=y)) = P(E=x)+P(E=y)

(OR can mean just one happens, or both happen (so it's not the same as XOR, "exclusive OR")).
(note that this only works for two possible outcomes of a single event)
So the probability that the first coin is heads OR the first coin is tails is 1/2 + 1/2 = 1.

Now, the formula for OR when dealing with two distinct events is more complicated. It utilizes something known as "the principle of inclusion-exclusion". To wit,

P((E=x) OR (F=y)) = P(E=x) + P(F=y) - P((E=x) AND (F=y)) = P(E=x) + P(F=y) - P(E=x)*P(F=y)

So, let's use these principles to figure your problem out. First, the problem says that the probability of any particular ring being returned is 10%, e.g. 1/10. That is, P(Ring = returned) = 1/10, and P(Ring = not returned) = 1 - P(Ring = returned) = 1 - 1/10 = 9/10.

So now, they are telling us that there were 5 rings sold. Let us name these Ring_1, Ring_2, Ring_3, Ring_4, and Ring_5.

They ask us what the probability that none of them are returned is. In other words, they want to know

P((Ring_1 = not returned) AND (Ring_2 = not returned) AND (Ring_3 = not returned) AND (Ring_4 = not returned) AND (Ring_5 = not returned))

By our formula, this is equal to

P(Ring_1 = not returned)*P(Ring_2 = not returned)*P(Ring_3 = not returned)*P(Ring_4 = not returned)*P(Ring_5 = not returned) = 9/10*9/10*9/10*9/10*9/10 = 59049/100000

Next, they ask what is the probability that at least 3 rings will be returned. Well, that means the desired outcomes are that 3 rings were returned OR 4 rings were returned OR 5 rings were returned.

P(3 rings were returned) OR (4 rings were returned) OR (5 rings were returned)) = P(3 rings were returned)*P(4 rings were returned)*P(5 rings were returned)

Next, notice that if three were returned, there are 10 ways this could happen. we could have rings 1,2,3 returned; or rings 2,3,4; or 3,4,5; etc. If you count all the ways you can pick 3 rings out of 5 rings, it turns out to be 10 ways. Thus, P(3 rings were returned) = 10*P(rings 1,2, and 3 were returned; 4,5 not returned) = 10*P(Ring_1 returned)*P(Ring_2 returned)*P(Ring_3 returned)*P(Ring_4 returned)*P(Ring_5 returned) = 10*(1/10*1/10*1/10*9/10*9/10) = 10*81/100000 = 81/10000
There are 5 ways to return exactly 4 rings (1,2,3,4; 1,2,3,5; 1,2,4,5; 1,3,4,5; 2,3,4,5). The probability that exactly 4 rings were returned is thus 5*P(rings 1,2,3,4 returned; ring 5 not returned) = 5*(1/10*1/10*1/10*1/10*9/10)= 5*9/100000 = 9/20000
Finally, the probability that 5 rings are returned is P(rings 1,2,3,4, and 5 are returned) = 1/10*1/10*1/10*1/10*1/10 = 1/100000.

Thus, the probability of at least 3 rings being returned is 81/10000 + 9/20000 + 1/100000 = 856/100000.

Factorization Problem

(Click on image below to expand)