animallogicmathThere are several chickens and rabbits in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many chickens are there, and how many rabbits?

Let c be the number of chickens, and r be the number of rabbits.
r + c = 72
4r + 2c = 200
To solve the equations, we multiply the first by two, then subtract the second.
2r + 2c = 144
2r = 56
r = 28
c = 44
So there are 44 chickens and 28 rabbits in the cage.

animalcleanfunnyWhat is a sick crocodile?

An ail-ligator.

animalshortI can swim or walk for miles.
I’m big with thick, white hair.
I live up in the Arctic.
What I am?

Polar bear.

cleanmathshortDivide 110 into two parts so that one will be 150 percent of the other. What are the 2 numbers?

44 and 66.

logicmathshortCan you make 10 plus 4 = 2?

Yes. 10 o'clock + 4 hours = 2 o'clock.

funnylogicshort A hobo had just been kicked off the train by one of the bosses. As he made his way down a dusty side road, he noticed a saffron robed man sitting next to a campfire apparently deep in thought. A wonderful smelling stew was bubbling in a pot next to him. It had been a full day since the hobo's last meal, so he went over to the man and tapped him on the shoulder.
"I see by your robes that you are some kind of holy man," said the hobo.
The Zen Master turned to the hobo and said, "You speak the truth."
The hobo spoke, "I would sure like to try the stew you have on the campfire there; perhaps if I could tell you something to increase your wisdom, you will agree to share your meal."
The Zen Master turned to the hobo and said, "Please, you are welcome to share my meal because you have already increased my wisdom!"
What had the Zen Master learned from the hobo to increase his wisdom?

The Zen Master learned that he should find a more privace place to meditate if he doesn't want to be interrupted by every vagabond that happens by.

funnylogic "Welcome back to the show. Before the break, Mr Ixolite here made it to our grand finale! How do you feel Mr.Ix?"
"Nervous."
"Okay, now to win the star prize of one million pounds all you have to do is answer the following question in 90 seconds."
"Okay, I'm ready."
"Right. In 90 seconds name 100 words that do NOT contain the letter 'A'. Start the clock!"
Can you help?

One, two, three, four, five...one hundred! I just counted from 1 to 100 in ninety seconds (it is possible).

logicA woman who lived in Germany during World War II wanted to cross the German/Swiss border in order to escape Nazi pursuers. The bridge which she is to cross is a half mile across, over a large canyon. Every three minutes a guard comes out of his bunker and checks if anyone is on the bridge. If a person is caught trying to escape German side to the Swiss side they are shot. If caught crossing the other direction without papers they are sent back. She knows that it takes at least five minutes to cross the bridge, in which time the guard will see her crossing and shoot her. How does she get across?

She waits until the guard goes inside his hunt, and begins to walk across the bridge. She gets a little more than half way, turns around, and begins to walk toward the geman side once more. The guard comes out, sees that she has no papers, and sends her "back" to the swiss side.

crazyfunnyshortWhy are televisions attracted to people ?

Because, people turn them on.

logicmathprobabilityWhat is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?

Only 23 people need to be in the room.
Our first observation in solving this problem is the following:
(the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0
What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations).
With some simple re-arranging of the formula, we get:
the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday)
So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer.
The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far.
These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918
More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365)
We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%.
Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.