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Panopticon
08-04-2012, 01:56 AM
For those who like this kind of stuff.

The only rule is that the problems are to be solved by the participants only. That means that no sources or anything are allowed to solve any kind of problem.

If you think you have the solution, post it and try to reason/explain for it and how you came to the solution.

After one problem is solved, feel free to post another one. Unfortunately, the one who posts the problem should probably not solve it as they probably already know the answer.

I'll start off with this probability problem:


How probable is it that at least two persons in a room of 23 people share the same birthday?

Insuperable
08-04-2012, 02:24 AM
P=1-[(365-1)/365]*[(365-2)/365]....*[(365-22)/365]= you should get high probability me thinks

I put 1-[] because it says at least and at least two means only one overlap

Dacul
08-04-2012, 02:38 AM
^You need to take into account leap years also:
http://en.wikipedia.org/wiki/Leap_year

Sikeliot
08-04-2012, 02:39 AM
Can this include logic puzzles and riddles?

Panopticon
08-04-2012, 02:51 AM
^You need to take into account leap years also:
http://en.wikipedia.org/wiki/Leap_year

Leap years aren't part of the problem. It would be silly to include leap years anyway, it isn't likely that you're born on the 29th February, 1/4 the chance of any other day.


Can this include logic puzzles and riddles?

Yes - all kinds of problems are allowed.

Sikeliot
08-04-2012, 02:55 AM
Ok here's one.

There are six eggs in the basket. Six people each take one of the eggs. How can it be that one egg is left in the basket?

Pretan
08-04-2012, 02:58 AM
One of the people takes the basket with them.

Sikeliot
08-04-2012, 03:01 AM
Right.

Here's another.

A man went into a party and drank some of the punch. He then left early. Everyone at the party who drunk the punch subsequently died of poisoning. Why did the man not die?

Panopticon
08-04-2012, 03:02 AM
The punch was poisoned after he had drank it and left?

Piper
08-04-2012, 03:03 AM
he poisoned the rest? :eek:

Sikeliot
08-04-2012, 03:04 AM
The punch was poisoned after he had drank it and left?

There are two possible answers.

1) He drank the punch, poisoned it himself, and then left.
2) The poison was in the ice cubes, which were still frozen when he drank it.

So take your pick which you like better. :thumb001:

Sikeliot
08-04-2012, 03:07 AM
A man and a woman live peacefully in a house together. But one day the woman shoots her husband. Then she holds him under water for over five minutes. Finally, she hangs him, but ten minutes later they go out and enjoy a wonderful dinner. How can this be?

Dacul
08-04-2012, 03:07 AM
Leap years aren't part of the problem. It would be silly to include leap years anyway, it isn't likely that you're born on the 29th February, 1/4 the chance of any other day.



Yes - all kinds of problems are allowed.

Well if you want an exact result you need to take leap years also.
The probability that is someone is born on 29th of February is about 1/4*1/366

Panopticon
08-04-2012, 03:07 AM
There are two possible answers.

1) He drank the punch, poisoned it himself, and then left.
2) The poison was in the ice cubes, which were still frozen when he drank it.

So take your pick which you like better. :thumb001:

If it was someone's intention to kill all the people attending the party, he would have poured it as liquid instead of waiting for it to melt as ice. I don't think you freeze poison anyway. However, it wasn't necessarily the guy who drank the punch who did it; e.g. Jews did it.

Sikeliot
08-04-2012, 03:09 AM
If it was someone's intention to kill all the people attending the party, he would have poured it as liquid instead of waiting for it to melt as ice. I don't think you freeze poison anyway. However, it wasn't necessarily the guy who drank the punch who did it; e.g. Jews did it.

That's why another answer could be, the man drank the punch and then poured in the poison himself to kill everyone else.

Pretan
08-04-2012, 03:09 AM
A man and a woman live peacefully in a house together. But one day the woman shoots her husband. Then she holds him under water for over five minutes. Finally, she hangs him, but ten minutes later they go out and enjoy a wonderful dinner. How can this be?

Well, what you have to ask yourself is in what context did she "Shoot" her husband.(Obviously not with a gun)

Piper
08-04-2012, 03:10 AM
A man and a woman live peacefully in a house together. But one day the woman shoots her husband. Then she holds him under water for over five minutes. Finally, she hangs him, but ten minutes later they go out and enjoy a wonderful dinner. How can this be?

the man = /= the husband

Sikeliot
08-04-2012, 03:13 AM
the man = /= the husband

Incorrect.

Pretan is on the right track.

Piper
08-04-2012, 03:14 AM
I'm not buying into the kinky sex theory, it's way too extreme and no one is into everything at once (bloodplay, axphysia etc)

I still think the man and the husband in the story are not the same person.

Sikeliot
08-04-2012, 03:15 AM
I'm not buying into the kinky sex theory, it's way too extreme and no one is into everything at once (bloodplay, axphysia etc)

I still think the man and the husband in the story are not the same person.


They are the same person. But she didn't "shoot" him with a gun. That's the hint.

Loki
08-04-2012, 03:15 AM
There are two possible answers.

1) He drank the punch, poisoned it himself, and then left.
2) The poison was in the ice cubes, which were still frozen when he drank it.

So take your pick which you like better. :thumb001:

... or 3) He didn't drink enough :p

Panopticon
08-04-2012, 03:16 AM
A man and a woman live peacefully in a house together. But one day the woman shoots her husband. Then she holds him under water for over five minutes. Finally, she hangs him, but ten minutes later they go out and enjoy a wonderful dinner. How can this be?

It's not meant to be taken literally. Metaphors for an argument or something?


Well if you want an exact result you need to take leap years also.
The probability that is someone is born on 29th of February is about 1/4*1/366

It simply complicates it too much and is unnecessary. It wouldn't be 1/366 anyway, that leap year only comes every 4th year. Which again makes it simply unnecessary to count leap years as well.


That's why another answer could be, the man drank the punch and then poured in the poison himself to kill everyone else.

I was just joking.

Sikeliot
08-04-2012, 03:17 AM
If you want me to give the answer to the one about the woman shooting the husband, I will :p

Pretan
08-04-2012, 03:17 AM
They are the same person. But she didn't "shoot" him with a gun. That's the hint.

She shot him with a professional camera and then developed the photo underwater, after she left it to dry.

Sikeliot
08-04-2012, 03:17 AM
She shot him with a professional camera and then developed the photo underwater, after she left it to dry.

Correct.;)

Piper
08-04-2012, 03:18 AM
She shot him with a professional camera and then developed the photo underwater, after she left it to dry.

freaking English:picard1:
yeah, you're totally right!

Sikeliot
08-04-2012, 03:22 AM
Three of these statements are untrue,

Mr Red: "Mr Blue did it."

Mr Blue: "Mr Red did it."

Mr Green: "Mr Blue's telling the truth."

Mr Yellow: "Mr Green's not lying."

Do you know who did it??

Piper
08-04-2012, 03:25 AM
mr Blue

Panopticon
08-04-2012, 03:26 AM
This one is relatively known.

You are at a game show. You have to pick between three doors, behind two of them there are goats, and behind one of them there is a car. You pick door 3. The show host opens door 1 that has a goat, and asks you if you want to change the door. Should you change door? And what are the probabilities now?

Panopticon
08-04-2012, 03:27 AM
Three of these statements are untrue,

Mr Red: "Mr Blue did it."

Mr Blue: "Mr Red did it."

Mr Green: "Mr Blue's telling the truth."

Mr Yellow: "Mr Green's not lying."

Do you know who did it??

Mr. Red. Do mine if anyone answered correct.

Sikeliot
08-04-2012, 03:27 AM
This one is relatively known.

You are at a game show. You have to pick between three doors, behind two of them there are goats, and behind one of them there is a car. You pick door 3. The show host opens door 1 that has a goat, and asks you if you want to change the door. Should you change door? And what are the probabilities now?

50/50.
Since there are 2 goats and 1 car and you now see that the door opened (which you did NOT pick) has a goat behind it, there is a 50% chance that the door you did pick is either a car or a goat, thus there is no obvious advantage to changing doors.

Sikeliot
08-04-2012, 03:28 AM
and the answer to mine was Mr. Blue. Since the other two corroborated his statement and 3 of the statements were untrue, Mr. Blue was the liar.

Panopticon
08-04-2012, 03:30 AM
50/50.
Since there are 2 goats and 1 car and you now see that the door opened (which you did NOT pick) has a goat behind it, there is a 50% chance that the door you did pick is either a car or a goat, thus there is no obvious advantage to changing doors.

I'll give you a hint: you had already picked a door when there were three possibilities.

Sikeliot
08-04-2012, 03:31 AM
I think I got it right.. I didn't?

Panopticon
08-04-2012, 03:32 AM
I think I got it right.. I didn't?

Nope. It's not that easy, lots of math professors in the US were wrong about it too when the question was asked.

Oh, and I missed the 3 of them are lying part in your previous puzzle.

Piper
08-04-2012, 03:32 AM
I'll give you a hint: you had already picked a door when there were three possibilities.

Triton already answered and I am of the same opinion: 50/50 chance of getting the car, no point in changing the choice. So what's the answer?

Panopticon
08-04-2012, 03:34 AM
Triton already answered and I am of the same opinion: 50/50 chance of getting the car, no point in changing the choice. So what's the answer?

Be patient. I want more people to answer, this is a problem that has given tons of math professors and such headaches.

Sikeliot
08-04-2012, 03:35 AM
I looked up the answer to the one before and I fail to see how switching your choice is beneficial.

Stefan
08-04-2012, 03:36 AM
This one is relatively known.

You are at a game show. You have to pick between three doors, behind two of them there are goats, and behind one of them there is a car. You pick door 3. The show host opens door 1 that has a goat, and asks you if you want to change the door. Should you change door? And what are the probabilities now?

Only thing I can think of is that it doesn't specify whether or not both a goat and a car can be behind the same door. So the chance is still 1/3rd?

Piper
08-04-2012, 03:39 AM
Only thing I can think of is that it doesn't specify whether or not both a goat and a car can be behind the same door. So the chance is still 1/3rd?

I guess it doesn't matter how many goats are, it states that behind one door there is a car. Surely you can dispose of the extra goat if you win the car, no?:D


still 2 door left and 50/50 chance IMO

Sikeliot
08-04-2012, 03:39 AM
The answer now makes sense to me now that I figured it out.

Panopticon
08-04-2012, 03:40 AM
Only thing I can think of is that it doesn't specify whether or not both a goat and a car can be behind the same door. So the chance is still 1/3rd?

They are evenly distributed between the doors, there are two doors with a goat behind, and one door with a car behind it.

I'll give you this as a hint, pretty bad, but it doesn't matter. You are on to something if you:

sWM4-Eg4GT8

Sikeliot
08-04-2012, 03:42 AM
Horten, I want to hear how you explain the answer.

Piper
08-04-2012, 03:43 AM
yeah, explain :eek:

Panopticon
08-04-2012, 03:46 AM
Horten, I want to hear how you explain the answer.

When you first chose a door, there was a 1/3 possibility that it was a car and a 2/3 possibility that there was a goat behind the door. You pick a door, 2/3 P that you picked the wrong door. The game show host decided to do you a favour by showing you one wrong door and giving you the option to change the door. The probability before he showed the door still matters because you have already picked a door, that door was 2/3 likely to be wrong. Since one wrong option has been eliminated the chance that the other door is the right one is 2/3, because you were 2/3 likely to be wrong.

Don't know if I explained it well.

Sikeliot
08-04-2012, 03:48 AM
But why would switching doors increase your chances if you don't know what's behind either of the remaining two doors? The point is that switching doors when asked increases your chance.

Stefan
08-04-2012, 03:50 AM
I understand the probability behind it, but does this actually work, experimentally? If we emulated this experiment again and again, and one person knew about this and the other person didn't, would that first person get more cars than the second? I'm trying to conceptualize it.

Sikeliot
08-04-2012, 03:51 AM
I don't understand why switching doors increases your probability. Why would it matter which you started with if the third door was a goat?

Piper
08-04-2012, 03:52 AM
When you first chose a door, there was a 1/3 possibility that it was a car and a 2/3 possibility that there was a goat behind the door. You pick a door, 2/3 P that you picked the wrong door. The game show host decided to do you a favour by showing you one wrong door and giving you the option to change the door. The probability before he showed the door still matters because you have already picked a door, that door was 2/3 likely to be wrong. Since one wrong option has been eliminated the chance that the other door is the right one is 2/3, because you were 2/3 likely to be wrong.

Don't know if I explained it well.

I get it, you explained well. I would still keep my choice if I'd be in the situation tho. Probability is just that, probability

Stefan
08-04-2012, 03:53 AM
I don't understand why switching doors increases your probability. Why would it matter which you started with if the third door was a goat?

You start out with a 1/3rd chance to get a car not knowing that the first door had a goat.

After you've gained that knowledge, it is clear that there is a greater chance you were wrong than right, and therefore the second door is the more probable choice. At least that's how I understand it.

Panopticon
08-04-2012, 03:55 AM
But why would switching doors increase your chances if you don't know what's behind either of the remaining two doors? The point is that switching doors when asked increases your chance.

You have chosen one door, the game show host doesn't open the door you had picked: he opens another door. So you are already stationed at door 3, with only door 2 as an option now; you can change or stay. Your choice of picking door 3 was 66% likely to be wrong. Therefore, if you switch, you are 66% likely to take the right door.

Remember that the possibilities when there were 3 doors is still there, because you have already picked one door before the show host revealed one door, but have the option to change it now. The chance would have been 1/2 if there had been only 2 options from the beginning.

Sikeliot
08-04-2012, 03:58 AM
You have chosen one door, the game show host doesn't open the door you had picked: he opens another door. So you are already stationed at door 3, with only door 2 as an option now; you can change or stay. Your choice of picking door 3 was 66% likely to be wrong. Therefore, if you switch, you are 66% likely to take the same door.

Remember that the possibilities when there were 3 doors is still there, because you have already picked one door before the show host revealed one door, but have the option to change it now. The chance would have been 1/2 if there had been only 2 options from the beginning.

I'm confused. If I pick door 1 and door 3 has a goat.. that's one goat already out of the picture and of the two remaining doors, one of them must be a goat and the other a car. Therefore why is it not a 50/50 chance given what I'd now know?

You're right if the question is "what was the chance you picked the right door from the start?" It'd be 1/3. but if door 3 is now eliminated, I fail to see how you have a 1/3 or 2/3 chance when you only now have to debate between two doors, one of which is a goat and the other a car.

Panopticon
08-04-2012, 03:59 AM
You start out with a 1/3rd chance to get a car not knowing that the first door had a goat.

After you've gained that knowledge, it is clear that there is a greater chance you were wrong than right, and therefore the second door is the more probable choice. At least that's how I understand it.

Basically - yes.


I get it, you explained well. I would still keep my choice if I'd be in the situation tho. Probability is just that, probability

This has actually been tested out in practice by a lot of schools, universities, etc. They found out that you should switch if you ever were to come in a situation like this.

Stefan
08-04-2012, 04:03 AM
I'm confused. If I pick door 1 and door 3 has a goat.. that's one goat already out of the picture and of the two remaining doors, one of them must be a goat and the other a car. Therefore why is it not a 50/50 chance given what I'd now know?

You're right if the question is "what was the chance you picked the right door from the start?" It'd be 1/3. but if door 3 is now eliminated, I fail to see how you have a 1/3 or 2/3 chance when you only now have to debate between two doors, one of which is a goat and the other a car.

It's because when you initially chose the third door, you had no idea the goat was behind the first. If there were only two doors from the start, the chance would be 50/50. However, you start out with a 33% chance to guess correctly. Chances are against you, not for you: and you made your choice when the chances were against you. Therefore since you know that the first door is a goat. The door you chose is 66% likely to be wrong, instead of 50% likely. So you'd rather go with the 66% door number 2 than stay with the original 33% chance of door number 3.

Sikeliot
08-04-2012, 04:04 AM
Stefan I am confused.. I thought you pick door 1, door 3 is opened for you, and you are asked to switch to door 2. Explain it with that in mind so I can get it. :lol:

Sikeliot
08-04-2012, 04:05 AM
I don't think it matters what the original odds were once one of the doors is eliminated, since you're now picking one out of TWO doors not three. You had a 1/3 chance of getting it right when you had to pick one of three doors. Now that the host opened one of them, you can count that door out and now just look at two.

I refuse to accept it's 1/3 and 2/3 chances.

Panopticon
08-04-2012, 04:06 AM
I'm confused. If I pick door 1 and door 3 has a goat.. that's one goat already out of the picture and of the two remaining doors, one of them must be a goat and the other a car. Therefore why is it not a 50/50 chance given what I'd now know?

You're right if the question is "what was the chance you picked the right door from the start?" It'd be 1/3. but if door 3 is now eliminated, I fail to see how you have a 1/3 or 2/3 chance when you only now have to debate between two doors, one of which is a goat and the other a car.

I'll go over it again: there are three doors, two with a goat, one with a car; you pick door nr. 1, this choice was 66% likely to be wrong, there are still three doors at this moment; the show host shows you that door nr. 3 has a goat behind it, but the chance that you were wrong on your initial pick of door nr. 1 is still 66%, it doesn't change; you know now that there is only one other option, but the choice you made at first was 66% likely to be wrong or 33% likely to be correct; therefore it's 66% likely that door nr.2 is the right one.

Panopticon
08-04-2012, 04:07 AM
I don't think it matters what the original odds were once one of the doors is eliminated, since you're now picking one out of TWO doors not three. You had a 1/3 chance of getting it right when you had to pick one of three doors. Now that the host opened one of them, you can count that door out and now just look at two.

I refuse to accept it's 1/3 and 2/3 chances.

But you have already chosen a door when the chance was 1/3 for picking the door with a car. This choice either remains or is changed after the show host decides that he will show you one of the doors that were incorrect.

Stefan
08-04-2012, 04:08 AM
Stefan I am confused.. I thought you pick door 1, door 3 is opened for you, and you are asked to switch to door 2. Explain it with that in mind so I can get it. :lol:

Basically you start out with a 1/3rd chance. You are given the choice again, after one of the other two doors is revealed. If you don't change your original choice, you still have that same probability of 1/3rd, because you originally started out with a 1/3rd chance and if you remain with the same door you also keep that 1/3rd chance. These choices aren't independent, but dependent of each-other. If you do change your choice, you have a 2/3rd probability, because it was not your original choice.

Sikeliot
08-04-2012, 04:08 AM
But when they open door 3, and you see it's a goat.. you are now picking between TWO doors. not three. You can count door 3 out like it was never there. It becomes "Of the two remaining doors, one is a goat one is a car." Thus why would it matter about the initial number of doors if you're essentially eliminating one of them?

Piper
08-04-2012, 04:10 AM
so how about you choose again...the door no. 3? :D

Sikeliot
08-04-2012, 04:11 AM
Why does it matter what your original chance was if door 3 is now basically eliminated and gone? To me you can now completely disregard that door ever having been there. since you're now left with two and you know one of them is a goat and the other a car.

Stefan
08-04-2012, 04:11 AM
But when they open door 3, and you see it's a goat.. you are now picking between TWO doors. not three. You can count door 3 out like it was never there. It becomes "Of the two remaining doors, one is a goat one is a car." Thus why would it matter about the initial number of doors if you're essentially eliminating one of them?

Because you know you were more likely to be wrong the first time, when you had three choices. So the chances are in favor that you were wrong, and the other door is the right one.

Sikeliot
08-04-2012, 04:12 AM
Well, I don't see why it matters that there were three doors to start if you can count out one of them altogether now that it's been revealed.. I look at it as, door #3 is now incorrect and there is one goat left, so you can pretend there was never a third door at all.

Sikeliot
08-04-2012, 04:15 AM
Try this..

http://www.theproblemsite.com/games/monty_hall_game.asp

I'll do it and just see if switching is beneficial.

Piper
08-04-2012, 04:15 AM
after all this talk I'm starting to think a goat wouldn't be such a bad of a prize:D

Stefan
08-04-2012, 04:15 AM
Well, I don't see why it matters that there were three doors to start if you can count out one of them altogether now that it's been revealed.. I look at it as, door #3 is now incorrect and there is one goat left, so you can pretend there was never a third door at all.

How about it like this.

Door #1


Door #2


Door #3


Combined door number 1 and 2 have a 2/3rd chance.

Door number #3 has a 1/3rd chance.

You choose door number 3.

Door number 1 is revealed, AFTER you chose door number 3.

That 2/3rds move to door number 2, because you made that choice of door number 3, WITHOUT the knowledge that door number one had a goat. If you knew BEFORE you chose door number 3, the chance would be 50/50.

Piper
08-04-2012, 04:17 AM
I got the concept, it's been explained very clearly. Now, hand me that goat please.:bored:

Sikeliot
08-04-2012, 04:18 AM
Ok I get it now. To understand it you have to consider your original odds, I guess.

Piper
08-04-2012, 04:19 AM
Try this..

http://www.theproblemsite.com/games/monty_hall_game.asp

I'll do it and just see if switching is beneficial.

I did it LOL
Replicated the whole thing. Chose 3, No 1 was empty, still chose 3 and I won

I didn't try more after that, how many cars one needs?:D

Panopticon
08-04-2012, 04:22 AM
I got the concept, it's been explained very clearly. Now, hand me that goat please.:bored:

It had to be taken into custody. Sorry. Anyone that comes from a mountainous areas aren't allowed to own goats. Something about 'vile mountaineer practices'. :noidea: You will have to settle with the car, I'm afraid.

Sikeliot
08-04-2012, 04:25 AM
I get it if I don't look at it as the third door having been eliminated, and that there are still 3 doors. My problem was acting like the third door was no longer there.

Loki
08-04-2012, 11:19 AM
Good thread, keep em coming :thumb001:

Stefan
08-09-2012, 09:32 AM
This is a very, very, easy one, but I want to bump the thread.

What is the smallest positive number you can make with a 1 and three 9s, using any operations?

Panopticon
08-09-2012, 09:36 AM
This is a very, very, easy one, but I want to bump the thread.

What is the smallest positive number you can make with a 1 and three 9s, using any basic operations?

9/9/9+1=1.11? Or 9+9/9-1=1.

Stefan
08-09-2012, 09:40 AM
9/9/9+1=1.11?

I don't know why my source said "basic operations" considering the nature of the answer. I hope that gives you a hint. ;)


Or 9+9/9-1=1.

Not this either

Minesweeper
08-09-2012, 10:02 AM
1/9*9*9= 1/729≈0,001

Stefan
08-09-2012, 10:14 AM
1/9*9*9= 1/729≈0,001

Closer, but not it. :)

Stefan
08-10-2012, 12:34 AM
Well it's 1/(9^9^9). :P Sorry about putting "basic operations", I pretty much copy and pasted the source and didn't realize it until afterwards.

Insuperable
08-13-2012, 01:29 AM
Bump


John, Mark and Luke are gathering plastic bottles. They earned half a dollar per bottle and they gathered 33 bottles. ( expensive I know but it is not important ). John gathered two times more than Mark and three times more than Luke. How many bottles each of them gathered?

Stefan
08-13-2012, 01:39 AM
Luke = 6

John = 18

Mark = 9

I solved it algebraically.

Luke = X

John = 3X

Mark = 3x/2

33 = x + 3x + 3x/2

66= 2x +6x + 3x

66 = 11x

x=6

Then I put 6 back in for x.

kabeiros
08-13-2012, 01:40 AM
John, Mark and Luke are gathering plastic bottles. They earned half a dollar per bottle and they gathered 33 bottles. ( expensive I know but it is not important ). John gathered two times more than Mark and three times more than Luke. How many bottles each of them gathered?Did you mean 33 dollars (which would translate to 66 bottles)? If so the answer is that John gathered 33 bottles, Mark gathered 22 bottles and Luke gathered 11 bottles. But if you were not wrong on that part, than John gathered 16.5 bottles, Mark gathered 11 and Luke gathered 5.5 (which makes no sense, how could they gather o.5 bottles)?

kabeiros
08-13-2012, 01:42 AM
Sorry my mistake, I thought you said that Mark gathered two times more than Luke...

Insuperable
08-13-2012, 01:42 AM
Did you mean 33 dollars (which would translate to 66 bottles)? If so the answer is that John gathered 33 bottles, Mark gathered 22 bottles and Luke gathered 11 bottles. But if you were not wrong on that part, than John gathered 16.5 bottles, Mark gathered 11 and Luke gathered 5.5 (which makes no sense, how could they gather o.5 bottles)?

Yes I meant 33 kunas not bottles
However the answer is not 33, 22, 11

Insuperable
08-13-2012, 01:49 AM
Luke = 6

John = 18

Mark = 9

I solved it algebraically.

Luke = X

John = 3X

Mark = 3x/2

33 = x + 3x + 3x/2

66= 2x +6x + 3x

66 = 11x

x=6

Then I put 6 back in for x.

Correct ( I did made a huge grammar word mistake but it is not important )

Stefan
08-13-2012, 01:59 AM
A simple and very easy logic problem. I always found these fun when studying for standardized tests.


The statement “If Jill misses the bus, she will be late” is true. Which other statement must be true?

(A) If Jill does not miss the bus, she will not be late.
(B) If Jill is not late, she missed the bus.
(C) If a student misses the bus, he or she will be late.
(D) Jill is late because she missed the bus.
(E) If Jill is not late, she did not miss the bus.

kabeiros
08-13-2012, 02:06 AM
(A) If Jill does not miss the bus, she will not be late. False, because she could be late even if she does not miss the bus
(B) If Jill is not late, she missed the bus. False, no explanation needed :)
(C) If a student misses the bus, he or she will be late. False, what the fuck are students doing in here? Jill is not a student, she works to the factory :D
(D) Jill is late because she missed the bus. True, but she could be late even if she didn't
(E) If Jill is not late, she did not miss the bus. True, the only one which is 100% true

Stefan
08-13-2012, 02:08 AM
^ Correct, indeed! The easiest way is to find the proof by contrapositive (http://en.wikipedia.org/wiki/Proof_by_contrapositive), but you can always just think through it, that's why I found them fun.

Stefan
08-13-2012, 02:28 AM
If v = (2,-3,5) m/s, what is the magnitude of 3v?

Insuperable
08-13-2012, 03:03 AM
v=2i-3j+5k m/s ---> 3v=6i-9j+15k m/s

modul=|3v| =sqrt(6^2 + 9^2 + 15^2)

Stefan
08-13-2012, 03:10 AM
Yep! 3 √38 simplified

Stefan
08-13-2012, 10:17 AM
Wolves are sometimes observed to have black coats and blue eyes. Assume that these traits are controlled by single locus genes and are located on different chromosomes. Assume further that normal coat color (N) is dominant to black (n) and brown eyes (B) are dominant to blue (b). Suppose the alpha male and alpha female of a pack (these are the dominant individuals who do most of the breeding) are black with blue eyes and normal colored with brown eyes, respectively. The female is also heterozygous for both traits. How many of the offspring (assume 16) living in the pack will have each of the following genotypes?

NNBB:
NNBb:
NNbb:
NnBB:
NnBb:
Nnbb:
nnBB:
nnBb:
nnbb:

Stefan
08-14-2012, 08:23 PM
Answer


NNBB:0
NNBb: 0
NNbb: 0
NnBB: 0
NnBb: 4
Nnbb:4
nnBB:0
nnBb:4
nnbb: 4

Female = heterozygous for both traits.

NnBb

Male = homozygous recessive for both traits.

nnbb

Either do two mono-crosses and apply the multiplication rule, or make a di-hybrid cross and count.

I did the first, since it's easier.

You get 1/2 Nn * 1/2 Bb = 1/4 NnBb

1/2 nn * 1/2 bb = 1/4 nnbb

1/2 Nn * 1/2 bb = 1/4 Nnbb

1/2 nn * 1/2 Bb = 1/4 nnBb

Stefan
08-15-2012, 09:00 AM
How many ways can you order the letters in the word "apricity"?

Insuperable
08-15-2012, 09:05 AM
8!/2!=3x4x5x6x7x8 ways

Stefan
08-15-2012, 09:16 AM
8!/2!=3x4x5x6x7x8 ways

Correct, I was confused before with the extra *2!. I was trying to figure out if I did this problem wrong, lol. ;)

It comes out to 20,160.