Well usually I am good with math but I have not taken statistics so I'm not very remembering of probability. This is not homework but a general question to get an idea of it. There are 10 positions for a position in the White House and suppose that there are only 3 elligible people to fill in the positions. So they all share the 10 spots but crazy enough 1 person can take them all Now how many possible combinations can there be?
.-. ehhh removed him from my friend list and was not sure if he changed name or not. I was hoping he would come on and answer his old username Magical Unicorn was way better than his current
No clue about the mathsbut...His/The person your talking abouts old name was Magical Unicorn? Idk who he is but sorry whoever you are I copied your name but didn't mean to, thought it was original (My alt's '-Magical_Unicorn-)
The probability of given 10 choose 3 with replacement it is 1000 without it is 720... The way you find it is the probability the first gets it is 10.. Second is 9... Third is 8 then multiply all by each other and you get your answer (non replacement) with you do 10*10*10... I'm in ap statistics now lol
I'm taking at least one statistics class next semester and two if I get my waitlist. The answer to your question is 3^10. To approach this question, lets say that there are 10 seats and 3 different people that can fill any given seat. Therefore, for the first seat there are 3 choices, for the second seat their are 3 choices, and so there are 3 choices for every seat. Then, by the Basic Principle of Counting, there are 10 experiments where there are 3 possible outcomes for each experiment which gives us 3^10 possible different arrangements. The general formula for this since each seat is independent of all of the other seats is p^n where n is the number of seats and p is the number of people.
Oh ok I understand now see I just hate the way some problems are worded that's my weakness in math lol