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All the information needed to solve the problem is there. The rate that they break down is uniform over 15 years (in other words, there is no day where it is more likely to break than any other day).
I hope this answers your question.
Dr. Pepper-
What are the odds that each machine will break down at least once in the fifteen-year period?
Does it make any difference?
Also (as I think “WolfishMusings” is asking), is there a limit on how many times a machine can break down during the fifteen years?
Once they break that is it, they are not fixed or replaced.
Dr. Pepper-
3) Years one and fifteen have an average 1.5-year daily window; day one, year one has only until the first day of year two, while the last day of year one has until the last day of year two. The average therefore is 1.5 years.
4) 15 years divided by a two-year window is 7.5. 15 years divided by 1.5 is 10.
5) 13 * 7.5 + 2 * 10 = 117.5. 117.5 / 15 = 7.83. Therefore, the odds of two 100% certainties each of which occurs one within fifteen years both occurring with a year of each other within that fifteen year span is 1/7.83, or about 12.8%.
If you’d said “within the same year”, it would be simply be 1/15 * the odds of both breaking.
I can only try-
Sorry, but you started off wrong, a uniform rate over fifteen years means that it can not survive past fifteen years. The odds of both machines breaking down in the fifteen year period is 100%.
Dr. Pepper-
In that case I’ll go with 1/7.83, or about 12.8%.
I can only try-
That’s the answer the I got. Here’s how I did it:
For the two machines, let’s call them x and y, plot their lives on a 15 * 15 graph with the life of x on the x-axis and the life of y on the y-axis. Draw a 45 degree angle line from the origin to 15,15 to show the possible places where x can break down. (It’s a straight line since any point in time has an equal probability of the machine breaking down there.)
Now draw two more 45 degree angle lines, the first should be shifted one unit to the left of the first line and the other one should be shifted one unit to the right of the first one. The area created by those two lines and bounded by the 15 * 15 graph is 1-196/225 = 12.88889%.
The way I calculated the area is by first calculating the “negative” area and subtracting. On the outside there are two triangles with the same area. Each is a right triangle with legs of length 14. the area is (14 * 14)/2, therefore the are of both triangles is 14 * 14 = 196. The total area of the graph is 15 * 15 = 225. Subtracting the triangles from the graph gives us 1 – (196/225) = 12.888889%.
Do you want to publish a paper on this with me? You can earn an Erd?s number of 5.
Dr. Pepper-
Neat solution – basically you’re removing a diagonal strip the width of one cube’s diagonal from the lower left corner to the upper right corner of a 15 cube x 15 cube square.
I have no idea how you even come up with this (and this is probably an easier problem for math meivinim).
Do you want to publish a paper on this with me? You can earn an Erd?s number of 5.
Well, thanks to Google I now know what an Erd?s number is.
As far as your offer, usually an Aleph Bina is not published in the same volume as a sefer Kabala ?.
The “Company” You Keep
1) We build excitement.
2) The quicker picker-upper.
5) Fly the friendly skies.
6) Hold the pickles, hold the lettuce.
7) The pause that refreshes.
8) We try harder.
9) Reach for the star.
10) Be all that you can be.
11) Hey, you never know.
17) Four out of five dentists recommend sugarless gum for their patients who chew gum.
18) When it rains, it pours.
19) We bring good things to life.
20) The one that coats is the only one you need.
I can only try-
We discussed once (I’m too lazy to link it) that equations to a mathematician are like tools for a plumber. When you are shown a solution it looks easy but without training you wouldn’t know which tools to use.
This, you said, was why plumbers charge so much- they have the experience to look at a problem and fix it. So, in essence you are paying for his experience and not his work.
As far as earning an Erd?s number of 5- there was a rumor that Hank Aaron earned an Erd?s number of 1- by signing a baseball with Paul Erd?s.
Dr. Pepper-
As far as earning an Erd?s number of 5- there was a rumor that Hank Aaron earned an Erd?s number of 1- by signing a baseball with Paul Erd?s.
The “Company” You Keep – Answers
1) We build excitement. Pontiac
2) The quicker picker-upper. Bounty Paper Towels
5) Fly the friendly skies. United Air Lines
6) Hold the pickles, hold the lettuce. Burger King
7) The pause that refreshes. Coca Cola
8) We try harder. Avis Rent-A-Car
9) Reach for the star. Texaco
10) Be all that you can be. United States Army
11) Hey, you never know. New York Lottery
17) Four out of five dentists [surveyed would] recommend sugarless gum for their patients who chew gum. Trident
18) When it rains, it pours. Morton Salt
19) We bring good things to life. General Electric (GE in the ad)
20) The one that coats is the only one you need. Pepto-Bismol
(clearly this was waaaay too easy to pique the interest of trivia aficionados here)
This doesn’t belong here but I didn’t think it warranted it’s own thread.
Did anyone try the Rubik’s 360?
I enjoyed solving it but I was disappointed that there was no strategy involved like the Rubik’s Cube or Rubik’s Clock.
just last week my nephew showed it to me. if it was mine i would have dropped a large brick on it and said: “solve THIS, rubik!”
Moderator-80-
It is kind of frustrating in the beginning since the center sphere has a different axis of rotation than the middle sphere but once you get the hang of it it becomes really easy.
There is also some fine motor skills needed but it gets easier in each subsequent attempt.
Finally! I just finished reading through the entire thread.
Hi, everybody I’m new to the CR. I actually just registered so I can try and stump you guys (i.e. Dr. Pepper).
Here goes.
You have 12 balls that look and feel identical but one of them is a slightly different weight. You don’t know if it is heavier or lighter. By using a balance (not a scale) only three times, how can you tell which one is different?
P.S. This is my all time favorite riddle.
If you read through the entire thread you should have come across that riddle already (the original one was with coins instead of balls).
I did read through all the posts, it took me almost a week. I’m 99% sure that this wasn’t posted yet. If I remember correctly that one was where you know that the odd one weighs more. If you read the riddle carefully you will see that it’s different. Remember you don’t know if the odd one weighs more or less. Let me know if I did actually miss it.
Good Luck
You’re probably correct.
(I’m not about to read through all the posts.)
Do you want me to post the solution?
If you can… heh heh heh…
coins 1,2,3&4 on one side 5,6,7&8 on other side
if balanced these 8 are even.
take 3 good coins on one side, take coin 9,10,&11 on other side
if balanced,coin 12 is bad, compare to any coin to see if heavy or lighter!
if 9,10&11 are not even and weigh more or less, put 9 on one side of scale and 10 on other side of scale, if even 11 is bad, if not then you know if it’s 9 or 10
if 1,2,3&4 goes to one side then 9,10,11&12 are good
take coin 1,5&6 on one side and coins 2,7&8 on other
if even 3 or 4 bad weigh against each other
if it goes down on one side we know one coin is light on side a or 1 of 2 is heavy on side b
weigh 2 coins that might be heavy if even 3rd coin is light if not which ever is heavy is the bad coin
“if 9,10&11 are not even and weigh more or less, put 9 on one side of scale and 10 on other side of scale … then you know if it’s 9 or 10”
chesedname – you still wouldn’t know if it 9 or 10, since you don’t know if the bad coin weighs more or less than the good coins.
Coins 9,10&11 on one side and 3 good coins on other if 9,10&11 weigh more than 3 good coins the coin we’re looking for is heavier and vice versa
That’s pretty impressive if you actually sat down and worked it out. In any case for clarity’s sake I’m copying and pasting the answer from a riddle website:
For the first weighing let us put on the left pan balls 1,2,3,4 and on the right pan balls 5,6,7,8. There are two possibilities. Either they balance, or they don’t. If they balance, then the different ball is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different ball is either 11 or 12. Weigh ball 1 against 11. If they balance, the different ball is number 12. If they do not balance, then 11 is the different ball. If 1,2 vs 9,10 do not balance, then the different ball is either 9 or 10. Again, weigh 1 against 9. If they balance, the different ball is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these balls could be the different ball. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different ball is either 3 or 4. Weigh 4 against 9, a known good ball. If they balance then the different ball is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy ball, or 1 is a different, light ball. For the third weighing, weigh 7 against 8. Whichever side is heavy is the different ball. If they balance, then 1 is the different ball. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy ball or 2 is a light different ball. Weigh 5 against 6. The heavier one is the different ball. If they balance, then 2 is a different light ball.
Interestingly enough, the answer that chesedname gave would provide us with the knowledge of whether the odd ball is heavier or lighter in every case.
There is a second solution to this riddle. See: http://256.com/gray/teasers/twelve_balls.html
Is there a a finite number of prime numbers? Prove that there is or isn’t.
There are an infinite number of prime numbers. If you take a bunch of prime numbers and multiply them by each other and add 1 you will get a prime number (2*3+1=7, 2*3*5+1=31). So therefore if, theoretically, there was a finite amount of primes (p1, p2, p3,… pn), and you multiplied them all by each other and added one (p1*p2*p3…*pn+1) you would get a new one so therefore it must be infinite.
YW Moderator-42-
In most cases you will not get a prime number when multiplying a few primes and adding 1. The exception is when one of the primes is 2, otherwise if you multiply a few primes and add 1 you will get an even number.
Other than that the proof is correct.
How many people have to be in a group in order for the chance that two of them have the same birthday be greater than 50%?
(For simplicity assume that all years have 365 days.)
No one? Hint… the answer has already been posted on this site. Now you just have to look through 3 years of my old posts 😀
squeak-
You may have been here since January of 07 but the CR hasn’t been around that long. Did you post the answer on the main page?
I did notice on an old post of yours that you were going to write an introduction about me when I publish my book. How’s that coming along?
It will be ready exactly 3 days after I am presented with the proofs.
Dr. Pepper
Member
In most cases you will not get a prime number when multiplying a few primes and adding 1. The exception is when one of the primes is 2, otherwise if you multiply a few primes and add 1 you will get an even number.
Am I to understand this as saying that 2 * any prime + 1 is a prime number?
Is 63 really a prime number? 63 = 1 + 2 * 31
Is this good enough proof?
http://www.theyeshivaworld.com/coffeeroom/topic/funny-shidduch-stories/page/13#post-74799
Proofs is what I said, and proofs is what I meant. Not proof.
If you’re really going to publish (by which I mean book publishing, not the academic kind), you might want to become familiar with that term.
squeak-
Sorry if I worded it wrong.
In most cases you will not get a prime number when multiplying a few primes and adding 1. If the number 2 is not one of the primes then there is no chance of (product of the primes) + 1 being a prime since it will be an even number and divisible by 2.
The exception to the above rule is when one of the primes is the number 2. In that case the (product of the primes) + 1 will be odd and the probability of it being prime is greater than 0% (but less than 100%).
Ahhh, that makes sense now. Thanks. Maybe if you had been my math teacher I would have become proficient in that elusive subject.
squeak-
I have no plans of publishing the book in the near future. I was trying to get you nervous.
Can you let me know where the hint is? I don’t recall this riddle being discussed before.
Thanks
It’s in the birthdays thread.
Writing does not make me nervous. I would gladly honor that promise, if and when the time comes.
Thanks, I must have missed it since I didn’t check it on a regular basis while it was active.
squeak-
I found the answer there but it’s a little cryptic. It looks as if there was a post deleted (possibly from “ames”).
You’re right. Probably a deleted post or two besides hers.
all of ames posts were deleted at her request.
its a long story
squeak:
Dr. Pepper
Member
In most cases you will not get a prime number when multiplying a few primes and adding 1. The exception is when one of the primes is 2, otherwise if you multiply a few primes and add 1 you will get an even number.
Am I to understand this as saying that 2 * any prime + 1 is a prime number?
Is 63 really a prime number? 63 = 1 + 2 * 31
What I meant is if you multiply all the primes in order from p1 to pn and add one you will get a prime.
YW Moderator-42-
The proof is correct but a different statement,
“If you take a bunch of prime numbers and multiply them by each other and add 1 you will get a prime number (2*3+1=7, 2*3*5+1=31)”
is not always correct as explained above.
Answer to the birthday riddle from last Tuesday:
If there are 23 people in the room there is a greater than 50% chance that two of them have the same birthday (as Squeak hinted at last year “If 23 people walk into a room…”).
For those interested, here’s how it can be solved in Excel:
People Odds Against Odds in Favor
1 =365/365 =1-B2
2 =B2*((365-(A3-1))/365) =1-B3
3 =B3*((365-(A4-1))/365) =1-B4
4 =B4*((365-(A5-1))/365) =1-B5
5 =B5*((365-(A6-1))/365) =1-B6
6 =B6*((365-(A7-1))/365) =1-B7
7 =B7*((365-(A8-1))/365) =1-B8
8 =B8*((365-(A9-1))/365) =1-B9
9 =B9*((365-(A10-1))/365) =1-B10
10 =B10*((365-(A11-1))/365) =1-B11
11 =B11*((365-(A12-1))/365) =1-B12
12 =B12*((365-(A13-1))/365) =1-B13
13 =B13*((365-(A14-1))/365) =1-B14
14 =B14*((365-(A15-1))/365) =1-B15
15 =B15*((365-(A16-1))/365) =1-B16
16 =B16*((365-(A17-1))/365) =1-B17
17 =B17*((365-(A18-1))/365) =1-B18
18 =B18*((365-(A19-1))/365) =1-B19
19 =B19*((365-(A20-1))/365) =1-B20
20 =B20*((365-(A21-1))/365) =1-B21
21 =B21*((365-(A22-1))/365) =1-B22
22 =B22*((365-(A23-1))/365) =1-B23
23 =B23*((365-(A24-1))/365) =1-B24
24 =B24*((365-(A25-1))/365) =1-B25
25 =B25*((365-(A26-1))/365) =1-B26
26 =B26*((365-(A27-1))/365) =1-B27
27 =B27*((365-(A28-1))/365) =1-B28
28 =B28*((365-(A29-1))/365) =1-B29
29 =B29*((365-(A30-1))/365) =1-B30
30 =B30*((365-(A31-1))/365) =1-B31Once there are 23 people the odds are greater than 50%. Interestingly, if there are 57 people the odss are greater than 99%.
When my oldest went to sleep last night she had twice as many sisters as brothers. When she woke up this morning she had the same amount of brothers and sisters.
What is the most likely explanation?
must be im reading this wrong. whats the riddle?
her mother had a baby boy during the night
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