Math extra credit?

An experiment consistes of choosing with replacement an integer at random among the numbers 1 to 19 inclusive. If we let M denote a number that is an integral multiple of 3 and N denote a number that is not an integral multiple of 3, arrange in order of increasing liklihood the following sequences of results:





a) MNNMN


b) NMMN


c) NMMNM


d) NNMN


e) MNMM

Math extra credit?
Between 1 and 19 inclusively, there are 6 multiples of three and thirteen numbers which are not multiples of three, so the probability of choosing a multiple of three=P(M)=6/19 and the probability of choosing a number that is not a multiple of three is P(N)=13/19.


The probability of a sequence of choices with replacement is the product of the probabilities of each choice made in the sequence.


Thus





P(d)=((13/19)^3) (6/19)





P(b)=((13/19)^2)((6/19)^2)


=(19/13)(6/19)(P(d))


=(6/13)(P(d)


%26lt;P(d) since 6/13%26lt;1





P(a)=((13/19)^3)((6/19)^2)


=(13/19)(P(b)


%26lt;P(b) since 13/19 %26lt;1





P(e)=(13/19)((6/19)^3)


=((19/13)^2)(6/19)P(a)


=((6)(19)/(13)^2)P(a)


=(114/169)P(a)


%26lt;P(a) since 114/169%26lt;1





P(c)=((13/19)^2)((6/19)^3)


=(13/19)P(e)


%26lt;P(e) since 13/19%26lt;1





So the sequences in order of increasing likelihood are c,e,a,b,d.
Reply:c)
Reply:do your homeowrk yourself.