You are not logged in.
Indeed, if you are looking at one car owner, he will be on his "6 Year car" even after 8 years !
But we are looking at a large population here, I imagine, who are changing their cars every day.
Possible future for Large Car Owners (Year 0,3 and 6):
LLL
LLS
LSL
LSS
The question is for just one car owner, though. But it does not say how long the owner has had his present car.
So, he/she may exchange his car later today! Or not for 3 years.
Just for Fun, I worked out that matrix "P" where
P^3 = [0.6 0.25]
[0.4 0.75]
It is (approximately):
[0.818 0.114]
[0.182 0.886]
So, what is P^6 ?
[0.460 0.338]
[0.540 0.662]
And P^8 is:
[0.422 0.362]
[0.578 0.638]
And P^9 is
[0.411 0.369]
[0.589 0.631]
There is some drift due to calculation accuracy, but if you worked P out more accurately you may have something workable. But there may well be a rigorous way to do this rather than my "hey, lets use Excel and see what we get" approach
thanks!
OK, well, we still have your "8 year" problem ...
... you could cheat and work out the probabilites at 6 and 9 and ratio in between.
Or, you could work out an equivalent matrix that works on 1 year intervals.
In other words, what is the Matrix "P" where:
P^3 = [0.6 0.25]
[0.4 0.75]
(This is just an off-the-cuff idea, may not be rigorous)
Hi, maths_buff.
Markov chains are not my specialty, I am hoping that Milos or one of the other members is better versed in these than I am.
Having said that, I think your probability matrix needs to be transposed:
L S
S = L [ 0.6 0.25 ]
S [ 0.4 0.75 ]
So, if someone owns a large car at time 0 we will have
x(0) = [1]
[0]
at time 1 (3 years hence):
x(1) = [ 0.6 0.25 ] [1] = [0.6]
[ 0.4 0.75 ] [0] [0.4]
at time 2 (6 years hence):
x(2) = [ 0.6 0.25 ]^2 [1] = [ 0.46 0.3375 ] [1] = [0.46]
[ 0.4 0.75 ] [0] [ 0.54 0.6625 ] [0] [0.54]
(Which has the values you had already mentioned)
Faaar Out!
You showed it!
Big Smile
Oooohh .. a PUZZLE !
(6*6+6)/6 = 7 ...no
(6+6)/6 + 6 = 8 ...no
We can help you - can you be more specific?
Thank you so much! That makes me so happy to read.
cool_me, you are genius ...
... I think
LOL !!
Vote:
The Mad Mathematicians Muddlings
Dark Discussions
Fun Forum
The Hangout
Cafe Infinity
The Set of All Topics
Go on, draw the lines, then!
Just as soon as they wake up!
The guys who posed it said "yes you can use the main signs add subtract multiply, divide not to sure weather too treat it as a logic problem?"
It seems to me that "Logic Problem" is a clue. It may not be solvable, I don't know.
answer = tin/cow
Is that right? LOL !!
Do I get a prize for answering it, then?
Perhaps if we used some Algebra on it
Start With: 1+1=x
Subtract x from both sides: 1+1-x=x-x
Subtract 1 from both sides: 1+1-x-1=x-x-1
Simplify: 1-x=-1
I think that is as far as I can simplify it.
"Think Outside The Square!"
And thank YOU very much for taking the time to write such a nice note.
He must be a good teacher to have such a nice recommendation.
Rats, you mean it isn't solved?
Well, then, we shall try harder ...
We can ALL help you, rivali. Just ask a question!
Like this? No, sorry.