POTD 62 + Other Stuff

POTD

Pretty nice geezer. The texturing is nice, and that's some pretty high poly modeling.

Yay, another double UMTYMP assignment. I just love 'em.

Whatever, anyways for those of you doing Mathcounts I'm going to try and give training tips via my blog. If any of you aren't doing Mathcounts and don't want to better yourselves, then promptly ignore everything below.

Probability:

Most of you probably can do pretty basic probability problems such as finding the probability of a coin flipping heads twice in a row. How do you do it? You simply multiply the probability of both events happening, so in the case of the coin flip we have (1/2) * (1/2) = 1/4. So far nothing too difficult, so lets up the complexity.

What is the probability that a ball starting in the blue oval will end up in the red oval, assuming the ball travels down and has an equal probability of moving down each chute?

Obviously the probability the ball ends up in the first oval beneath the blue oval is 1. Now the probability that the ball ends up in the oval above the red oval is 2/3, since the ball has three paths and two of them takes it to the oval we want it at. Now the ball has two choices: to go right and end up in the red oval, or go left. So our total probability is 1*(2/3)*(1/2) = 1/3.

As you might have guessed multiplying probabilities in not the only way find a probability.

Lets consider this problem: Albatross, Bobert, Caddlebot, Deberg, and Eustien are in the senate. Three of them will be randomly chosen to head a committee to study the causes of bureaucratic inefficiency. What is the probability that Albatross will be in the committee? Now the old way of calculating probabilities won't work too well. Another way is to take the total number of possibilities that work and divide by the number of possibilities total. So in the case of this problem we would count the number of committees that include Albatross and then divide by the total number of committees.

Let us calculate the number of committees total. Here we use an operation called "choose". Choose is denoted (b,a) except that b would be on top of a; it's like a vertical parenthesis. (b,a) tells use the number of committees of size a that we can choose from a group of b people. The formula is (b,a) = b!/(a!*(b-a)!), where ! denotes factorial. I think you know this, but just to recap 5! = 5*4*3*2*1 and 0! = 1.

So the total number of committees to form is (5,3) = 5!/(3!*2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10. Now we calculate the number of committees that include Albatross, at this point we could just count them, but that isn't that efficient. A better way is to assume that we have already chosen Albatross, and then we are picking the remaining two members of the committee. So the number of committees that include Albatross is (4,2), since we are choosing two people out of a group of four remaining people. This number is 6. So the probability is 6/10=3/5.

I'll leave you with a challenge problem. A bug is walking on a triangle. A "move" consists of the bug moving from the vertex he is standing on to one of the other two vertexes. After 10 moves what is the probability that the bug will return to the vertex it started on?
Hint: Consider the moves he makes as a series of letters, with R denoting a right turn, and L denoting a left turn.
Hint 2: Think about the relation between the number of R and L turns it makes to where it ends up, in particular think of the number of right moves it must make to return to the original point.

I'll post the solution after a while...

This is definitely the longest post I have ever made. Woohoo!