stumped (for the first time...)

[policecap]

well, I was going through the 2008 National Sprint round and i came across 2 problems that were fairly tricky for me (?).

27. Three players each have a red card, blue card, and green card. The players play a game that consists of three rounds. In each round, each player randomly turns over one of his/her cards without replacement. What is the probability that, at the end of the game, one card of each color was turned over in each of the three rounds?
Well, that's one boring game alright.

AND

22. To the right is an example of a square grid of dots with three dots along each side. In a similar square grid of dots with 8 dots along each side, how many distinct lines pass through the bottom left corner and at least
one other dot of the grid?

sorry, no pic

so anyone who can help me will receive three helping dollars cough cough

(which can be spent on virtually nothing)

[limac]

For #27, I think it is 1/18 (correct me if I am wrong.)

Here's the explanation (If that is right...)
Click to reveal hidden content

Let's take the first round first, the first person can flip over any of the cards (so his choice doesn't matter), the second person can only flip over 2 of the possible 3 cards, thus with probability 2/3. The third person has only one choice for the last card to flip, with probability 1/3. Thus the probability for a successful round 1 is 2/9.

For round 2 (each of the people have 2 cards left), the first person starting can pick any of the two (so his choice doesn't matter), the second person has only 1 out of the 2 that is going to yield success. Similarly, the last person also is going to pick the last remaining card (the color which hasn't been picked yet), with probability 1/2. Thus the probability for success in this round is 1/4.

In round three, each of the players have only one card left. Since both, rounds 1 and 2 have to be successful, we have to multiply the probabilities. So the answer is 2/9 * 1/4 = 1/18 (Basically because, for each of the 2/9 times round 1 is successful, round 2 is successful 1/4 of those times.)

[Thunder365]

#27

WOW I am stupid. I didnt read the part on how you dont replace your cards, so I just did (2/9)^3.

Yeah, thats right then Limac. You could do: (# of ways to succeed on round 1)/(# of total ways to do round 1)x(# of ways to succeed on round 2)/(# of total ways to do round 2)x(# of ways to succeed on round 3)/(# of total ways to do round 3).

Round 1 you win 3x2x1 times out of a total of 3x3x3 times
Round 2 you win 2x1x1 times out of a total of 2x2x2 times.
Round 3 you only have 1 card left for everyone, so if you succeeded on round 1 and 2, you will succeed on round 3, so its 1.

for a total of (6)/(27)x(2)/(8)x(1)/(1) =

#22

As for number 22, its not that hard if you just list the possible slopes systematically, take out the fractions that equals a fraction listed before it (ie. for 3/2 and 6/4 we would need to take out 6/4 since its the same slope) then add 2 (you will see why). The slopes are:
1/1 1/2 1/3 1/4 1/5 1/6 1/7
2/1 2/3 2/5 2/7
3/1 3/2 3/4 3/5 3/7
4/1 4/3 4/5 4/7
5/1 5/2 5/3 5/4 5/6 5/7
6/1 6/5 6/7
7/1 7/2 7/3 7/4 7/5 7/6

But then we need to add 2 because we didnt list the 2 where the slope is undefined (follows y=7 when plotted one a grid) or when the slope is 0.

S0 there are a total of 35+2=

[mathlearner2012]

You are not allowed to post answers. If you want to find the answer, there is another site.

[Thunder365]

[mathlearner2012]

Wow I have like ten posts that tells me AIME15 said this rule. And your right, I am fairly new.