math extra credit?

Question

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

Answer 1

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)

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

P(e)=(13/19)((6/19)^3)
=((19/13)^2)(6/19)P(a)
=((6)(19)/(13)^2)P(a)
=(114/169)P(a)

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

So the sequences in order of increasing likelihood are c,e,a,b,d.

Answer 2

do your homeowrk yourself.

Answer 3

c)