The worlds first riddle!

[size=7]one minus five on six times five on six[/size]

Arriving almost on time
Pondering this furiously while watching the end of the world as we know it on the Discovery Channel
[size=7]Set it up like this, break everything into 10 minute windows
I (Rap) walked in at 5:10 & to succeed Markr came in at 5:00 or 5:10, failure is anything else---2 out of 6
Rap walked in at 5:20 & to succeed Markr came in at 5:10 or 5:20, failure is anything else---2 out of 6
Rap walked in at 5:30 & to succeed Markr came in at 5:20 or 5:30, failure is anything else---2 out of 6
Rap walked in at 5:40 & to succeed Markr came in at 5:30 or 5:40, failure is anything else---2 out of 6
Rap walked in at 5:50 & to succeed Markr came in at 5:40 or 5:50, failure is anything else---2 out of 6
Rap walked in at 6:00 & to succeed Markr came in at 5:50, failure is anything else----1 out of 6
0-10 min 2 out of 6
10-20 min 2 out of 6
20-30 min 2 out of 6
30-40 min 2 out of 6
40-50 min 2 out of 6
50-60 min 1 out of 6
add all together
0-60min (1 hour) 11 out of 36

Probability of Rap and Markr arriving within 10 minutes of each other is 11/36[/size]

I can live with that answer

Rap

Arriving almost on time
Pondering this furiously while watching the end of the world as we know it on the Discovery Channel
[size=7]Set it up like this, break everything into 10 minute windows
I (Rap) walked in at 5:10 & to succeed Markr came in at 5:00 or 5:10, failure is anything else---2 out of 6
Rap walked in at 5:20 & to succeed Markr came in at 5:10 or 5:20, failure is anything else---2 out of 6
Rap walked in at 5:30 & to succeed Markr came in at 5:20 or 5:30, failure is anything else---2 out of 6
Rap walked in at 5:40 & to succeed Markr came in at 5:30 or 5:40, failure is anything else---2 out of 6
Rap walked in at 5:50 & to succeed Markr came in at 5:40 or 5:50, failure is anything else---2 out of 6
Rap walked in at 6:00 & to succeed Markr came in at 5:50, failure is anything else----1 out of 6
0-10 min 2 out of 6
10-20 min 2 out of 6
20-30 min 2 out of 6
30-40 min 2 out of 6
40-50 min 2 out of 6
50-60 min 1 out of 6
add all together
0-60min (1 hour) 11 out of 36

Probability of Rap and Markr arriving within 10 minutes of each other is 11/36[/size]

I can live with that answer

Rap

RESTAURANT


[size=7]The square (ignore aspect ratio) represents arrival times for Rap and Mark. They arrive within 10 minutes of each other when their arrival times fall in the diagonal band. The combined area of the two triangles is (5/6)^2. Therefore, the area of the diagonal band is 1-(5/6)^2 = 11/36.
[/size]

Slippy:

one minus five on six times five on six



Rap:

Arriving almost on time
Pondering this furiously while watching the end of the world as we know it on the Discovery Channel

Set it up like this, break everything into 10 minute windows
I (Rap) walked in at 5:10 & to succeed Markr came in at 5:00 or 5:10, failure is anything else---2 out of 6
Rap walked in at 5:20 & to succeed Markr came in at 5:10 or 5:20, failure is anything else---2 out of 6
Rap walked in at 5:30 & to succeed Markr came in at 5:20 or 5:30, failure is anything else---2 out of 6
Rap walked in at 5:40 & to succeed Markr came in at 5:30 or 5:40, failure is anything else---2 out of 6
Rap walked in at 5:50 & to succeed Markr came in at 5:40 or 5:50, failure is anything else---2 out of 6
Rap walked in at 6:00 & to succeed Markr came in at 5:50, failure is anything else----1 out of 6
0-10 min 2 out of 6
10-20 min 2 out of 6
20-30 min 2 out of 6
30-40 min 2 out of 6
40-50 min 2 out of 6
50-60 min 1 out of 6
add all together
0-60min (1 hour) 11 out of 36

Probability of Rap and Markr arriving within 10 minutes of each other is 11/36

I can live with that answer


Ok! But what happened to the end of the world as we know it?



Mark:

The square (ignore aspect ratio) represents arrival times for Rap and Mark. They arrive within 10 minutes of each other when their arrival times fall in the diagonal band. The combined area of the two triangles is (5/6)^2. Therefore, the area of the diagonal band is 1-(5/6)^2 = 11/36.





Draw a 6 by 6 square. Plot the arrival time of one person vertically. Plot the arrival time of the other person horizontally. The times they arrive within 10 minutes of each other can be represented by a diagonal stripe across the square. (As per Mark's diagram)

The area of the entire square is 36. The area of the portions outside the stripe is 25. Thus the area of the stripe is 11. Considering the uniformity of the arrival times the probability of arriving within ten minutes is the area of the stripe divided by the area of the square = 11/36 =~ 30.6%


How about that; we all got the same answer.





UTMES


TWIIMTHE

MINUTES

WITHIN TIME

End of the world as we know it---

within 200 years of the demise of the human species--gaia recovers balance. Within 500 years, it is almost as if we never existed except for the occasional stainless steel sink, aluminum can, and disposable diaper. Pigs, goats, cats, cows and dogs go feral and survive quite well as Darwin principles sorts the survivors. Mankind and civilization will quickly be shown to be an aberration, a failed experiment.

Amazing--a relative large volcanic eruption, or a moderate meteor strike will have a greater disturbance than the last 8000 years of humanities improvement of nature.

Personally though, I can hardly wait.

Rap

Lzzie:

UTMES = MINUTES


TWIIMTHE = WITHIN TIME

That took you all of five minutes; what kept you!



Thanks for the heads up Rap, I thought it was gonna be bad news! :wink:





DANGER! Calculus and knowledge of the exponential distribution required.

You have ten light bulbs. Five have an average life of 100 hours, and the other five have a average life of 200 hours.

These light bulbs have a memory less property in that their current age (measured in how long they have already been on) has no bearing on their future life expectancy.

Assuming they are all already on what is the expected number of hours before the first one burns out

Hint: The density function for this kind of light bulb with average life of n hours is f(x)=1/n * e^-x/n.

back in a moment i need to generate a function

Slippy, "back in a moment i need to generate a function"

I don't blame you, if they blow; just take them back under the guarantee.



asouunddible



bcoraerwd

sound inaudible

Good morning weekenders!

Well, it appears we had a little down time, I hope you spent it wisely.



TTH:

asouunddible = sound inaudible

Pardon!



This is what I would have posted yesterday:

10% of the people in a certain population use an illegal drug.

A drug test yields the correct result 90% of the time, whether the person uses drugs or not. A random person is forced to take the drug test and the result is positive.

What is the probability he uses drugs

DRUG TEST
[size=7]flip a fair coin[/size]

bcoraerwd?

inboard crew?

Mark:


DRUG TEST
flip a fair coin



The probability of event A given event B is Pr(A and B)/Pr(B).
In this specific case this is [(.1)(.9)]/[(.1)(.9)+(.9)(.1)] = 1/2.

Dang if he ain't right!


Lzzie:

bcoraerwd = inboard crew

I have: Crew inboard; same difference!





Simple question: What is the smallest integer greater than 0 that can be written entirely with zeros and ones and is evenly divisible by 225

225
[size=7]11100001 which is binary for 225. You didn't specify a base. [/size]

225
[size=7]11111111100 (base 10)
The number has to be a multiple of 900 (225*4). To be a multiple of 900, the sum of the digits must be a multiple of 9 and it must end with 00. 111111111 is the smallest number that is a multiple of 9 that is made up of ones and zeros. Add two zeros to the end, and you're done.[/size]

[size=7]11,111,111,100[/size]

Rap

Mark I:

225
11100001 which is binary for 225. You didn't specify a base.


Very true, now divide that by 225!



Mark II:

225
11111111100 (base 10)
The number has to be a multiple of 900 (225*4). To be a multiple of 900, the sum of the digits must be a multiple of 9 and it must end with 00. 111111111 is the smallest number that is a multiple of 9 that is made up of ones and zeros. Add two zeros to the end, and you're done.



Rap:

11,111,111,100





The prime factorization of 225 is 5*5*3*3. So the answer will be both a multiple of 25 and of 9.

All multiples of 25 end in either 00, 25, 50, or 75. The only one of these composed of 0's and 1's is obviously 00, so the answer must end in 00. The hard part is finding a series of 0's and 1's preceeding the 00 that will make the entire number divisible by 9.

If you didn't already know the following trick then this problem would be very hard. If you did know it then the problem was likely very easy. The trick is that if the sum of digits of a number is divisible by 9 then the number itself is also divisible by 9. Note that this is true for 3 also. For example the number 17685 is divisible by 9 because 1+7+6+8+5=27, and 27 is divisible by 9.

To prove this let's consider any five digit number, abcde. This number can be expressed as follows.

a*10000 + b*1000 + c*100 + d*10 + e =
a*(9999+1) + b*(999+1) + c*(99+1) + d*(9+1) + e*1 =
a*9999 + b*999 + c*99 + d*9 + a + b + c + d + e =
9*(a*1111 + b*111 + c*11 + d*1) + a + b + c + d + e

Thus 9*(a*1111 + b*111 + c*11 + d*1) is a multiple of 9. So if a+b+c+d+e is also a multiple of 9 then the entire number must be a multiple of 9. Note also that the remainder of abcde/9 is the same as the remainder of (a+b+c+d+e)/9.

The smallest number consisting of all 1's and divisible by 9 is thus 111,111,111. Adding the two zeros at the end results in the answer to the problem: 11,111,111,100.

But then, they already knew that!



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AZETTPPAECLK

oppinated

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