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Linear Algebra 16 Online
OpenStudy (anonymous):

On a given day the air quality in a certain city is either good or bad. Records show that when the air quality is good on one day, then there is a 95% chance that it will be good on the next day, and when the air quality is bad on one day, then there is a 45% chance that it will be bad on the next day.

OpenStudy (anonymous):

(b) If the air quality is good today, what is the probability that it will be good two days from now?

OpenStudy (anonymous):

i am not sure what i am doing wrong. question is related to markov chain rules

OpenStudy (b87lar):

I can help. Can you show what you did so far?

OpenStudy (anonymous):

i can type a bit... i have written in my rough copy so you cant understand what i did is

OpenStudy (anonymous):

.96G+0.55B=G eq 1 G+B=1 eq 2 i solve them by substitution

OpenStudy (anonymous):

my answer is 0.9168 which should be 0.93.

OpenStudy (b87lar):

well, I'd start with listing all the possible chains of events:

OpenStudy (b87lar):

starting with good air (G), going to next day, and then going to day after that

OpenStudy (anonymous):

i did by two method i use this foemula also (I-P)q=0

OpenStudy (b87lar):

I'd see these possibilities: G->G->G and G->B->G

OpenStudy (b87lar):

(we know we start with G and end with G)

OpenStudy (anonymous):

i did that also. i got 0.91672

OpenStudy (b87lar):

okay let's see, so the probability of the first GGG sequence is: 0.95*0.95

OpenStudy (anonymous):

can you just show me the starting step how would you do like equations?

OpenStudy (b87lar):

the prob of the other sequence GBG is 0.05*0.55

OpenStudy (anonymous):

its corrct

OpenStudy (b87lar):

sure i can show you the equations. Are you familiar with the conditional notation, like this P(G|G) ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

how we can solve this by markov chain rule?

OpenStudy (b87lar):

P(G|G) stands for the prob that, given Good air quality today, we will see Good tomorrow

OpenStudy (anonymous):

ok

OpenStudy (b87lar):

The Markov chain rule says that by multiplying P(G|G)*P(G|G) we get P(GG|G), which is the probability that, given Good today, we will get Good tomorrow AND Good day after that.

OpenStudy (b87lar):

so this allows us to just multiply the Markov probabilities together to get the probabilities of those sequences I was talking about

OpenStudy (b87lar):

so for GGG, I multiply P(G|G)*P(G|G) and for GBG multiply P(G|B)*P(B|G)

OpenStudy (b87lar):

those two sequences probabilities need to be added because they are both possible (one OR the other)

OpenStudy (b87lar):

and so we get 0.93

OpenStudy (anonymous):

can you make a tree diagram for me ? i got it now whats going on.

OpenStudy (b87lar):

|dw:1456634476568:dw| is that what you mean?

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