Howdy,

Calculators and interpreters (BASIC etc) are usually found on everyones computer; or can be easily added. In the old days when I was teaching BASIC (I even spoke with John Kemeny once) I found that a simulation ( random numbers or straight iteration) was often the easiest way to a solution. .

Luke should decide if that's OK by his rules. Since answers here are often required to several decimal places then calculators should be the order of the Day.

Others (such as me) may want to look up the solution to an integral that they have long forgotten. I guess I am in favor of using the computer to any advantage--even though it may access information compiled by others--kind of like the way of the world now.

kind of an interesting question for this thread--

What do you think luke, Kevvy, others ,,,??

best regards,

Bill
AKA DADDIO

That's a fair point... Bill.
Use them [Calculators + Scripts] when you have too.

Q16 is up for Grabs:

16. EASY (WORTH 1 POINT) - An urn contains a number of colored balls, with equal numbers of each color. Adding 20 balls of a new color to the urn would not change the probability of drawing (without replacement) two balls of the same color.
How many balls are in the urn? (Before the extra balls are added.)

(Mr. Kevvy and William Rothamel may NOT answer the above (No. 16) question....)

Best Regards,
Luke.

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- Luke.

*Bump*

Ive had one PM on this question. Just a note please answer the question in this thread....

Q16 is STILL up for Grabs:

16. EASY (WORTH 1 POINT) - An urn contains a number of colored balls, with equal numbers of each color. Adding 20 balls of a new color to the urn would not change the probability of drawing (without replacement) two balls of the same color.
How many balls are in the urn? (Before the extra balls are added.)

(Mr. Kevvy and William Rothamel may NOT answer the above (No. 16) question....)

Best Regards,
Luke

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- Luke.

Q16: EASY (WORTH 1 POINT) - An urn contains a number of colored balls, with equal numbers of each color. Adding 20 balls of a new color to the urn would not change the probability of drawing (without replacement) two balls of the same color.
How many balls are in the urn? (Before the extra balls are added.)

Q17: A baby is added to a hospital nursery. Before the baby was added there were two boys in the nursery and an uncounted number of girls. After the new baby is added a baby is selected at random among all the babys. The selected baby is a boy. What is the probability that the added baby was a girl?

Best Regards,
Luke.

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- Luke.

Is Q17 an "Easy" or "Hard"? ie, do Mr. Rothamel and myself get to answer it?

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You may answer all questions.... there are no limits for you two.... except for the fact that you two will now only be awarded 1/2 a point for every question....
The reason being: "You two are too good"

Best Regards,
Luke.

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- Luke.

190 balls. 19 colors, 10 of each.
P(drawing 2 of the same color) = 1/21.

After adding 20 of a new color,
P(drawing 2 of the same color) = (19*choose(10,2) + choose(20,2)) / choose(210,2), which is still 1/21.

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Capitalize on this good fortune, one word can bring you round ... changes.

190 balls. 19 colors, 10 of each.
P(drawing 2 of the same color) = 1/21.

After adding 20 of a new color,
P(drawing 2 of the same color) = (19*choose(10,2) + choose(20,2)) / choose(210,2), which is still 1/21.

Correct Sarge!
Answer: 190 balls. 19 colors, 10 each.

And the Stats are in....
1. Mr. Kevvy - 7 Points
2. WinterKnight - 4 Points
3. William Rothamel - 3 1/2 Points
4. John McLeod VII - 1 Point
5. Sarge - 1 Point
6. Scary Capitalist - 1/2 Point
7. TBD....

And Question 17 Remains.... (This is open to everyone)
Q17: A baby is added to a hospital nursery. Before the baby was added there were two boys in the nursery and an uncounted number of girls. After the new baby is added a baby is selected at random among all the babys. The selected baby is a boy. What is the probability that the added baby was a girl?

Best Regards,
Luke.

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- Luke.

THE ANSWER IS 40% OR 4 OUT OF 10

A good one Luke, however I must stipulate that without any information (pre-conditions) I am assuming equal probability that any birth is a 50-50 chance--which may not be exactly true-however for my answer to be correct I must assume this ratio.

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THE ANSWER IS 40% OR 4 OUT OF 10

A good one Luke, however I must stipulate that without any information (pre-conditions) I am assuming equal probability that any birth is a 50-50 chance--which may not be exactly true-however for my answer to be correct I must assume this ratio.

Correct!!! Well done, Bill!
Official Q17 Answer: 40%

Statistics Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 4 1/2 Points
3. WinterKnight - 4 Points
4. John McLeod VII - 1 Point
5. Sarge - 1 Point
6. Scary Capitalist - 1/2 Point
7. TBD....

Now something a bit different....
Normal questions will resume later.....

18 Beta: How many digits of Pi do you know? (Please be truthful)
(Open to everybody)
The question is worth 1/2 a point to the winner with the most memorized digits....

Best Regards,
Luke.

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- Luke.

THE ANSWER IS 40% OR 4 OUT OF 10

. . . but two out of three ain't bad... ;)

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It may not be 1984 but George Orwell sure did see the future . . .

THE ANSWER IS 40% OR 4 OUT OF 10

A good one Luke, however I must stipulate that without any information (pre-conditions) I am assuming equal probability that any birth is a 50-50 chance--which may not be exactly true-however for my answer to be correct I must assume this ratio.

Correct!!! Well done, Bill!
Official Q17 Answer: 40%

Statistics Table:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 4 1/2 Points
3. WinterKnight - 4 Points
4. John McLeod VII - 1 Point
5. Sarge - 1 Point
6. Scary Capitalist - 1/2 Point
7. TBD....

Now something a bit different....
Normal questions will resume later.....

18 Beta: How many digits of Pi do you know? (Please be truthful)
(Open to everybody)
The question is worth 1/2 a point to the winner with the most memorized digits....

Best Regards,
Luke.

3.14159265

I used to know it one place further out.
I suspect the next is a 4.
So, 9 or 10 places.

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Capitalize on this good fortune, one word can bring you round ... changes.

3.14159265

I used to know it one place further out.
I suspect the next is a 4.
So, 9 or 10 places.

LOL..I only remember as far as 3.141 ..but that is usually far enough for most purposes.

If Luke sets some questions that aren't probabilities I'll have a go. Probabilities are my worst part of maths. :(

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Reality Internet Personality

3.14159265

I used to know it one place further out.
I suspect the next is a 4.
So, 9 or 10 places.

LOL..I only remember as far as 3.141 ..but that is usually far enough for most purposes.

If Luke sets some questions that aren't probabilities I'll have a go. Probabilities are my worst part of maths. :(

Thats way too accurate, 3 is fine for me most of the time, calculating electronic components, but have to remember 3^2 = ~10. If I need more accurate its time for the plastic brain.

And I totally agree about the probability and statistics questions. Should be a completely separate subject taught in the business management and marketing building.

3.14159265

I used to know it one place further out.
I suspect the next is a 4.
So, 9 or 10 places.

LOL..I only remember as far as 3.141 ..but that is usually far enough for most purposes.

If Luke sets some questions that aren't probabilities I'll have a go. Probabilities are my worst part of maths. :(

Thats way too accurate, 3 is fine for me most of the time, calculating electronic components, but have to remember 3^2 = ~10. If I need more accurate its time for the plastic brain.

And I totally agree about the probability and statistics questions. Should be a completely separate subject taught in the business management and marketing building.

The degree of accuracy needed will, of course, vary.

No, probability and statistics is far from separate from mathematics.
Yes, business majors need to know some of it.
And so do math majors. To suggest otherwise means either the connections to geometry and calculus was never shown or forgotten.

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Capitalize on this good fortune, one word can bring you round ... changes.

3.14159265

I used to know it one place further out.
I suspect the next is a 4.
So, 9 or 10 places.

LOL..I only remember as far as 3.141 ..but that is usually far enough for most purposes.

If Luke sets some questions that aren't probabilities I'll have a go. Probabilities are my worst part of maths. :(

Thats way too accurate, 3 is fine for me most of the time, calculating electronic components, but have to remember 3^2 = ~10. If I need more accurate its time for the plastic brain.

And I totally agree about the probability and statistics questions. Should be a completely separate subject taught in the business management and marketing building.

The degree of accuracy needed will, of course, vary.

No, probability and statistics is far from separate from mathematics.
Yes, business majors need to know some of it.
And so do math majors. To suggest otherwise means either the connections to geometry and calculus was never shown or forgotten.

It's kind of important for Quantum Mechanics and Thermal Physics too.

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Reality Internet Personality

Howdy,

Ya Know, there is a sentence that you learn to remember 20 or so digits.. then there is another one to get the next 20 or so. I never did memorize the sentence though. It would make a good trick at a cocktail party or a Math Department Christmas party.

Poe, E.
Near a Raven

Midnights so dreary, tired and weary.
Silently pondering volumes extolling all by-now obsolete lore.
During my rather long nap—the weirdest tap!
An ominous vibrating sound disturbing my chamber's antedoor.
"This," I whispered quietly, "I ignore."

Perfectly, the intellect remembers: the ghostly fires, a glittering ember.
Inflamed by lightning's outbursts, windows cast penumbras upon this floor.
Sorrowful, as one mistreated, unhappy thoughts I heeded:
That inimitable lesson in elegance—Lenore—
Is delighting, exciting . . . nevermore.

That takes care of the first 80 or so .....

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ITI SAPIS
POTANDA
TINONE

Sarge wins with 9 digits... and by the way.... the next digit is a 3....

But if I could have a go myself... I would win the question..... this is gonna' get a lot of scowls..... I know off by heart 100 decimals of pi.... no more, no less..... also since I memorized them... my memory is a lot better....

100 Decimals of Pi: 3.1415926535897932384626433832795028841971693993751058209749445923078
164062862089986280348253421170679

Standings Table (Sponsored by Google Corp.)
1. Mr. Kevvy - 7 Points
2. William Rothamel - 4 1/2 Points
3. WinterKnight - 4 Points
4. Sarge - 1 1/2 Points
5. John McLeod VII - 1 Point
6. Scary Capitalist - 1/2 Point
7. TBD....

Es99 (part i) requested less probability questions.... I will do my best to try find some different ones for you!

But for now.... Have a go everyone on these....

18. Five pirates have come across a treasure of 1000 coins. According to pirate rules the pirate of highest rank must make a suggestion on how to divide the money. If a majority agree to his suggestion then it is to be followed by all the pirates. However, if the suggestion does not get a majority approval then the suggesting pirate is thrown overboard, after which time the remaining pirate of highest rank then makes a suggestion under the same rules. This process repeats, if necessary, until only the pirate of lowest rank is left, in which case he would get everything. Any pirate may suggest any distribution, rank does not guarantee getting more coins than anybody else. Assume that all pirates are infinitely greedy, infinitely logical, and infintely bloodthirsty, and that each pirate knows this to be true of every other pirate. The highest priority of each pirate is to get as much money for themselves as possible. The second highest priority is to throw overboard the other pirates. A pirate will vote to throw another one over even if they have no monetary gain by doing so, and even if it would cost them their own life, but would not if throwing them over would cost even 1 coin. How should the first pirate suggest dividing the money?

19. Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

Also with Q19....
There are 2 Hints.... PM me for one!

The first hint is "free" (In a sense that if you answer correctly you will be awarded the full point)
The Second Hint if requested by the same user halves your point if answered correctly.....

Have a go & Best Regards,
Luke.

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- Luke.

19. Three people (A, B, and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. There is also a bicycle available and any person can cross the bridge in 1 minute with the bicycle. What is the shortest time that all men can get across the bridge? Each man travels at their own constant rate.

A is the slowpoke, so give A the bicycle, he arrives at the end in 1 minute.
C arrives at the end in 2 minutes, takes the bicycle, rides back to B who is currently at the 2/5 mark so they have to cover 3/5 in total to intersect.
B covers 1/5 of the bridge per minute, C on bicycle covers 5/5 per minute, so they intersect in (3/5)/(1/5+5/5) = 1/2 minute. Time now 2 1/2 minutes.
B takes bicycle and goes to end; don't have to calculate that as C is now the bottleneck.
B was (2 1/2)/(5) or 1/2 of the way across the bridge leaving 1/2 left. C runs back to end in 1 minute and all have crossed.
Total time is 3 1/2 minutes or 3 mins. 30 seconds.

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A: 998 coins
B: 0 coins
C: 1 coin
D: 0 coins
E: 1 coin