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...?
The integers in question are 1,2,3,4,5,6,7,8,9. Of these, 3,6,9 are integer multiples of 3 and 1,2,4,5,7,8 are not integer multiples of three. Thus the probabilities are:





N (not a multiple of 3) 6/9 = 2/3


M (multiple of 3) 3/9 = 1/3


Check: a number either is an integer multiple of 3 or it isn't, so the cases are mutually exclusive and there are no other cases. 1/3 + 2/3 = 1 so these values of M and N are plausible.





Drawing is with replacement so the probability of the outcomes does not change during the experiment.





MNNMN = M^2 * N^3 = 1/9 * 8/27 = 8/243


NMMN = M^2 * N^2 = 1/9 * 4/9 = 4/81


NMMNM = M^3 * N^2 = 1/27 * 4/9 = 4/243


NNMN = M * N^3 = 1/3 * 8/27 = 8/81


MNMM = M^3 * N = 1/27 * 2/3 = 2/81





Now bring these all to a common denominator, 243





MNNMN = M^2 * N^3 = 1/9 * 8/27 = 8/243


NMMN = M^2 * N^2 = 1/9 * 4/9 = 12/243


NMMNM = M^3 * N^2 = 1/27 * 4/9 = 4/243


NNMN = M * N^3 = 1/3 * 8/27 = 8/81 = 24/243





Therefore the sequences arranged in order of increasing likelihood are


NMMNM, MNMM, MNNMN, NMMN, NNMN


and that's


c, e, a, b, d
Reply:c) NMMNM





I think
Reply:Let M denote a number that is an integral multiple of 3: 3, 6, or 9.





You don't say what N is, but I assume it is 1, 2, 4, 5, 7, or 8.





Probability of M, denoted P(M)=3/9=1/3.


Probability of N, denoted P(N)=6/9=2/3.





Probability of MNNMN


(1/3)(2/3)(2/3)(1/3)(2/3)


=8/243.





Probability of NMMN


(2/3)(1/3)(1/3)(2/3)=4/81


=12/243.





Probability of NNMN


(2/3)(2/3)(1/3)(2/3)=8/81


=24/243.





Probability of MNMM


(1/3)(2/3)(1/3)(1/3)= 2/81


= 6/243.





d is least likely.


a is more likely.


b is even more likely.


c is most likely.
Reply:??!!?!?


if M is an integral multiple of 3....


wat's N???