Math Geniuses!! Please help me with this Problem of the Week...?

I'd be passing Trig if i get this one! So please help me out as much as you can, and i'll love you forever :)


WARNING: THIS IS SUPER HARD. NO ONE CAN FIGURE IT OUT IN MY CLASS (and i have a smart class)





An experiment consists of choosing with replacement an integer at random among the numbers from 1 to 9, inclusive. If we let M denote a number that is an integral multiple of 3, arrange in order of increasing likelihood the following sequences of results:


a) MNNMN b) NMMN c) NMMNM d) NNMN e) MNMM





This is like a foreign language to me.


*I double-checked i made no errors in my typing so if it makes no sense, it not me, its my psychotic trig teacher. Thanks guys!!! Even if you dont get it, i give you credit for trying!!!

Math Geniuses!! Please help me with this Problem of the Week...?
Let's take this slow. We know that the numbers are not exclusive, so they will be available each time an integer is drawn. Now, there are three multiples of three between 1 and 9 (3, 6, and 9); the odds of getting one of those is 3 in 9, or 1/3. There are six other numbers, then (1,2,4,5,7, and 8); the odds of getting one of those is 6 in nine, or 2/3.





So,


a) The odds of this one is (1/3*2/3*2/3*1/3*2/3=6/243=2/81)


b) 2/3*1/3*1/3*2/3=4/81


c) 2/3*2/3*1/3*2/3=8/81


d) 1/3*2/3*1/3*1/3=2/81





You should be able to arrange them yourself. I hope that helps.
Reply:i got MNMM
Reply:We know that an "integral multiple of 3" would be 3, 6, and 9.





Since you are allowed to draw numbers with replacement, you have a 3/9 chance of drawing an integral multiple of 3. So, you have a 1/3 chance of drawing M.





For any string of x length, you can expect x/3 digits to be of the form M. In short, 33% of the string should be M. Of the strings you list, which is most likely? Count the Ms and divide by the length:


a - 2/5 = 40%


b - 2/4 = 50%


c - 3/5 = 60%


d - 1/4 = 25%


e - 3/4 = 75%





Of those, 40% is closest to 33% (though 25% is really, really close). So, I would guess the answer is a.





That's assuming I understand the question fully. If I'm wrong, I hope I at least gave you something to play with.





Edit: I missed the point about arranging them in order of likelihood. If you follow my method, then I'd go with a, d, b, c, e.


Although, when looking at the other answer, I am more inclined to go with the probability of each sequence happening where M = 1/3 and N = 2/3. I'd suggest his route.