Qfyilyi

I have the following problem:

• If today is a sunny day, a probability that it will rain tomorrow is $0.2$.


• If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.

I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.

My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.

A binary tree is definitely a possible way to solve this problem.

Another way to think about it though is maybe in the language or linear algebra.

We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.

So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.

More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.

edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$
$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}