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"Simple" Maths Problems - CLOSED!
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William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
it's obvious that it's 17 DADDIO |
Philadelphia Send message Joined: 12 Feb 07 Posts: 1590 Credit: 399,688 RAC: 0 |
Ok, instead of starting at 1 and working up I started at 100 and worked down, lol. I think it's 97 by trying it in my head, which now hurts. I could certainly be wrong though since there are so many other numbers to devide it by that I may have screwed up. [edit] oh, btw, this is not an easy one to do when you have to do it in your head. |
jason_gee Send message Joined: 24 Nov 06 Posts: 7489 Credit: 91,093,184 RAC: 0 |
Any Guesses???!!!! Come on.... this ain't even a maths problem!!! Well you must be thinking of 71, because it's written in the question. Anything else and you're thinking of two primes in that range ;D "Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Sorry Bill & Philly! jason gee got the answer! Well Done to you! Q71 Answer: 71! Microsoft Windows Standings 98 1. Fred W - 17 Points 2. William Rothamel - 16 1/2 Points 3. Guido.Man - 9 Points 4. Mr. Kevvy - 7 Points 5. Dominique - 6 1/2 Points 6. WinterKnight - 6 Points 7. Sarge - 5 1/2 Points 8. Philadelphia - 5 1/2 Points 9. John McLeod VII - 1 Point 10. Labbie - 1 Point 11. Jason gee - 1 Point 12. Scary Capitalist - 1/2 Point 13. TBD... Q72 (1 Point) : A woman is chosen at random among all women that have two children. She is asked do you have at least one boy, and she answers 'yes.' What is the probability her other child is a boy? Assume every pregnancy has a 50/50 chance to be a boy or a girl Luke. - Luke. |
Philadelphia Send message Joined: 12 Feb 07 Posts: 1590 Credit: 399,688 RAC: 0 |
Sorry Bill & Philly! The answer is 50%. It's like flipping a penny, the chances of it comming up heads is always 50/50 regardless of the previous results, in her case a boy. |
jason_gee Send message Joined: 24 Nov 06 Posts: 7489 Credit: 91,093,184 RAC: 0 |
... My Guess: [corrected: one third = 33.33%] chance of second child being a boy also. i.e. probabilities: boy-> boy = 25% boy-> girl = 25% girl-> boy = 25% girl->girl = 25% (left out of possibilities) Total = 100% [there are 3 possibilities with at least one boy, 2 of those remaining 3 possibilities have girls, leaving a one third chance of there being two boys] "Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
The answer is 2/3 or 66% that the other is a boy. Whoops--I see that the earlier answer appears to correct at 1/3 Bill |
Philadelphia Send message Joined: 12 Feb 07 Posts: 1590 Credit: 399,688 RAC: 0 |
... I would agree with that with the exception that we 'know' the first part of the problem, it was given to us as a 'boy', so the 3rd and 4th part of your situation can't happen as 'assumed'. Therefore, given a "M" first, the only two options left or either "M" or "F", thereore 50/50. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Well done to Jason Gee, Congratulations! 1 Point... Q72 Answer: 33% or 1/3 ALSO: IMPORTANT NOTICE!!! About 75 posts remain in this thread, before I announce a winner! So Everyone will have to hurry up if they want the Grand Prize! Prizes include: Thread Champion (1st Place) Thread Vice-Champion (2nd Place, Runner-up sounds sorta' lousy considering...) Late Entrant Winner (Any Participant who entered the comp later e.g. Philadelphia) Spot Prize (Any Participant who has points, drawn randomly) Spot Prize 2 (Any Participant who has points, drawn randomly) Everyone else will get congratulated and credited for participating and will happily welcomed to a "second thread" (Maybe...) Microsoft Windows Standings 98 SE 1. Fred W - 17 Points 2. William Rothamel - 16 1/2 Points 3. Guido.Man - 9 Points 4. Mr. Kevvy - 7 Points 5. Dominique - 6 1/2 Points 6. WinterKnight - 6 Points 7. Sarge - 5 1/2 Points 8. Philadelphia - 5 1/2 Points 9. Jason gee - 2 Points 10. John McLeod VII - 1 Point 11. Labbie - 1 Point 12. Scary Capitalist - 1/2 Point 13. TBD... Also - Bill, Keep on trying for top spot again, your only half-a-point away! Question 73 (1 Point) : Jack and Jill each have marble collections. The number in Jack's collection in a square number (1,4,9,16, etc). Jack says to Jill, "If you give me all your marbles I'll still have a square number." Jill replies, "If you gave me the number in my collection you would still left left with an even square." What is the least number of marbles Jack has? Luke. - Luke. |
Philadelphia Send message Joined: 12 Feb 07 Posts: 1590 Credit: 399,688 RAC: 0 |
Well done to Jason Gee, Congratulations! 1 Point... Congrats Jason. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Running through the first few squares it looks like Jack at 25 and Jill at 24 would do it. I presume that "EVEN SQUARE" does not mean the square of an EVEN number but means an "exact square". DADDIO |
jason_gee Send message Joined: 24 Nov 06 Posts: 7489 Credit: 91,093,184 RAC: 0 |
I'd have a go at this one too, but It's 3am Saturday morning here ... and I lost my marbles... "Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Congratulations to Bill! for solving Q73! Q73 Answer: Jack has 25 and Jill has 24 Q73 Solver: Bill And with this, Bill regains top position! Congrats again, Bill!. Where is Fred? Microsoft Windows Standings 2000 1. William Rothamel - 17 1/2 Points 2. Fred W - 17 Points 3. Guido.Man - 9 Points 4. Mr. Kevvy - 7 Points 5. Dominique - 6 1/2 Points 6. WinterKnight - 6 Points 7. Sarge - 5 1/2 Points 8. Philadelphia - 5 1/2 Points 9. Jason gee - 2 Points 10. John McLeod VII - 1 Point 11. Labbie - 1 Point 12. Scary Capitalist - 1/2 Point 13. TBD... Thanks Guido.man for trying! Here's another one: Question 74 (1 Point) : You are traveling down a path and come to a fork in the road. A sign lays fallen at the path indicating that one path leads to a village where everyone tells the truth and the other to a village where everyone tells lies. The sign has been knocked down so you do not know which path leads to which village. Then someone from one of the villages (you don't know which one) comes down the path from which you came. You may ask him one question to determine which path goes to which village. What question do you ask? (3 different answers exist) Luke. - Luke. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Congratulations to Bill! for solving Q73! Q74: "Would the road you have just come down take me to the village where all speak the truth?" If he answers "Yes" then assume that he leaves in the direction of the den of falsehoods, otherwise he will be heading towards the metropolis of virtue. F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
You ask him which is the way to his village. In this way you will always get directions to the truth teller's village and the other is to the liar's village. Go thru the logic and you will see that I am right. You must assume that the liars will ALWAYS tell a lie. I think Guido should get a point for the last one too. Daddio |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Both Answers by Fred & Bill seem to be correct, so I'll award a point to each of you, and as per Bills request, guido.man shall be given a point for Q73... Microsoft Windows Standings ME 1. William Rothamel - 18 1/2 Points 2. Fred W - 18 Points 3. Guido.Man - 10 Points 4. Mr. Kevvy - 7 Points 5. Dominique - 6 1/2 Points 6. WinterKnight - 6 Points 7. Sarge - 5 1/2 Points 8. Philadelphia - 5 1/2 Points 9. Jason gee - 2 Points 10. John McLeod VII - 1 Point 11. Labbie - 1 Point 12. Scary Capitalist - 1/2 Point 13. TBD... Congratulations to guido.man for being the third person to top 10 points! Q75 : 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? Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Hint on #75 you must use Bayes' Theorem. The answer will give you surprising results to see how unreliable this test would be. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
I was not totally happy with my answer to Q74 since it assumes the villager is going directly home and not just wandering round aimlessly or going to visit someone in the other village :) So here is a second attempt (after a night's sleep) "If I were to ask you, and I am not actually asking you, which of these 2 roads leads to village xxxx, what would you tell me?" The truthful villager would point the right way, the other, had you asked the question directly would have pointed the wrong way but now has to give the opposite answer. F. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Fred is right about his answer The answer to 75 is only 50%. Bayes' theorem is a little hard to understand. Extra puzzle for Fred for fun : If you came to a 5-way split in a road and saw that the signpost that pointed the way and mileage to all of the distant towns had been knocked over and was lying down--broken entirely from its moorings. How could you determine which road to take? |
jason_gee Send message Joined: 24 Nov 06 Posts: 7489 Credit: 91,093,184 RAC: 0 |
#75: a - probability that the person is a drug user is given as 10% b- probability the person is not a drug user is 1-0.1 = 90% c- probability the test is positive given the person is a drug user is 90% d- probability the test is positive given the person is NOT a drug user is 10% e- probability of a positive test is : (90% x 10%) + (10% x 90%) = 18% --> probability that the positive test indicates a drug user is : ( c x a) / e = (cxa)/((cxa)+(dxb))) = (.9x.1)/((.9x.1)+(.1x.9)) = .09 / .18 = ...50% ... Flip a coin , you're sacked ;D "Living by the wisdom of computer science doesn't sound so bad after all. And unlike most advice, it's backed up by proofs." -- Algorithms to live by: The computer science of human decisions. |
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