## Hat in the Red Sea

If you drop a yellow hat in the Red Sea, what does it become?

Wet.

If you drop a yellow hat in the Red Sea, what does it become?

Wet.

It was a grandeur party. In order to filter the uninvited guests, the security guard was assigned a task to check the secret password. The guests invited by the royal family also were shared with the secret password.
John wasn't an invited guest. He learned that the password is needed to make an entry. He hides himself and started watching the guests and the security.
The first guest comes. Security told him, TWELVE and the guest replied SIX. He wished him and allowed him to enter.
The second guest comes. Security told him SIX and the guest replied THREE! He was too allowed.
John made an entry as third guest. Security told him EIGHT and John replied FOUR. He was thrown out of the party!
Why?

The answer should be five. The password is not half of the digit, but the number that represents the number of digits told by security.

An iron horse with a flaxen tail.
The faster the horse runs,
the shorter his tail becomes.
What is it?

A needle and thread.

The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.
What is the most bananas you can bring over to your destination?

First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A.
So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again.
Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between.
When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward.
In the first part, P1, to shift the bananas by 1Km, the Camel will have to
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana
Note: After point 5 the Camel does not need to return to point A again.
So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas.
After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas.
Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km.
Move forward with 1000 bananas – Will eat up 1 banana in the way forward
Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back
Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Note: After point 3 the Camel does not need to return to the starting point of P2.
So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas.
After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas.
The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example.
Hence the length of part P2 is 333 Km.
Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas.
He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.

What 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.

You overhear a man talking to a clerk in a hardware store. The clerk says "One will cost you 12 cents, ten will cost your 24 cents, and one hundred will cost you 36 cents."
What is the man buying?

The man is buying physical numbers to nail to the front of his house. Each number costs 12 cents, and so "1" will cost 12 cents, "10" will cost 24 cents, and "100" will cost 36 cents.

I am a vehicle. I spell the same when you read me forwards as well as backwards. What Am I?

Race car.

I like to twirl my body but keep my head up high. After I go in, everything becomes tight. What am I?

Screw.

What did the banana do when it heard the ice scream?

It split.

I travel all over the world, but always stay in my corner. What am I?

A stamp.