An infinite number of mathematicians are standing behind a bar. The first asks the barman for half a pint of beer, the second for a quarter pint, the third an eighth, and so on. How many pints of beer will the barman need to fulfill all mathematicians' wishes?
There 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?
There are 44 chickens and 28 rabbits in the cage.
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
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.
A man who lives in Middletown has two girlfriends, one in Northtown and one in Southtown. Trains from the Middletown train station leave for Northtown once every hour. Separate trains from the station also leave for Southtown once every hour. No trains go to both Northtown and Southtown.
Each day he gets to the Middletown train station at a completely random time and gets onto the first train that is going to either Northtown or Southtown, whichever comes first.
After a few months, he realizes that he spends 80% of his days with his girlfriend from Northtown, and only 20% of his days with his girlfriend from Southtown.
How could this be?
The train to Northtown leaves every hour, on the hour (9:00AM, 10:00AM, etc...).
The train to Southtown leaves at 12 after the hour (9:12AM, 10:12AM, etc...).
So there is only a 12/60 (1/5) chance that he will end up on the train to Southtown each day, since he will usually get to the station during the 48 minutes of each hour when the train to Northtown will be the next to come.
Your friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it.
He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar.
You win if the marble you pick is blue, and you lose otherwise.
To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away).
How should you move the marbles around to give yourself the best chance of picking a blue marble?
Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.
Tarun Asthnaiya go to his office by local train. However nearby train station is quite far from his place and he used to drive his bike to train station daily with an average speed of 60km/hr. One day at halfway point he relized that due to heavy traffic he got late having average speed of just 30km/hr. How fast he must drive for the rest of the way to catch my local train?
The train is just about to leave the station and there is no way Tarun will be able to catch it this time.