Coding Style: define a function for each question. please use python$-3$

And plase make sure the code compiles this is my third time posting this question because the answer given do not compile. Use the latest version of all the libraries please and python $3\mathrm{.}11$

Given the Bayesian Networks Inference example discussed in the class, answer the following questions.

$1)$ The class example includes four random variables: Rain, Maintenance, Train and Appointment. Their dependency relations are: Maintenance depends on Rain, Train depends on Rain and Maintenance, Appointment depends on Train. Construct the Bayesian network and visualize it$.($Tip: you can use pgmpy$,$ pomegranate or any popular Python library to construct the Bayesian network, and use matplotlib or other library to visualize the network structure$).$

$2)$ Based on the constructed Bayesian network, make the following inferences: $(1)$ P $($Rain$=$heavy,

Maintenance$=$yes$),(2)$ P$($Appointment$=$attend $|$ Rain$=$none, Maintenance$=$no$)$

$2$

$3)$ Suppose the probability of attending or missing today$$s Appointment is not dependent on Train, but

only dependent on yesterday$$s Appointment. Given the initial state $[1,0]$ as the probability

distribution of $$attend$$ and $$miss$,$ and the transition matrix P $=[[0\mathrm{.}9,0\mathrm{.}1],[0\mathrm{.}5,0\mathrm{.}5]],$ what is the

probability distribution of $$attend$$ and $$miss$$ ten days later? $($Tip: Simulate the Markov chain over n

steps$).$

Conditional Probability Tables used for $1)$ and $2)$:

P$($Rain$)=($

"none": $0\mathrm{.}7,$

"light": $0\mathrm{.}2,$

"heavy": $0\mathrm{.}1$

$)$

P$($Maintenance$|$ Rain$)=([$

$["$none$",$ "yes", $0\mathrm{.}4],$

$["$none$","$no$",0\mathrm{.}6],$

$["$light$",$ "yes", $0\mathrm{.}2],$

$["$light$","$no$",0\mathrm{.}8],$

$["$heavy$",$ "yes", $0\mathrm{.}1],$

$["$heavy$","$no$",0\mathrm{.}9]$

$])$

P$($Train $|$ Rain, Maintenance$)=([$

$["$none$",$ "yes", $"$on time", $0\mathrm{.}8],$

$["$none$",$ "yes", "delayed", $0\mathrm{.}2],$

$["$none$","$no$","$on time", $0\mathrm{.}9],$

$["$none$","$no$",$ "delayed", $0\mathrm{.}1],$

$["$light$",$ "yes", $"$on time", $0\mathrm{.}6],$

$["$light$",$ "yes", "delayed", $0\mathrm{.}4],$

$["$light$","$no$","$on time", $0\mathrm{.}7],$

$["$light$","$no$",$ "delayed", $0\mathrm{.}3],$

$["$heavy$",$ "yes", $"$on time", $0\mathrm{.}4],$

$["$heavy$",$ "yes", "delayed", $0\mathrm{.}6],$

$["$heavy$","$no$","$on time", $0\mathrm{.}5],$

$["$heavy$","$no$",$ "delayed", $0\mathrm{.}5],$

$])$

P$($Appointment$)=([$

$["$on time", "attend", $0\mathrm{.}9],$

$["$on time", "miss", $0\mathrm{.}1],$

$["$delayed$",$ "attend", $0\mathrm{.}6],$

$["$delayed$",$ "miss", $0\mathrm{.}4]$

$])$

the transition matrix used for $3)$:

P $=$ $$0\mathrm{.}90\mathrm{.}1$

$0\mathrm{.}50\mathrm{.}5*$

It means P$($attend $->$ attend$)=0\mathrm{.}9,$ P$($attend $->$ miss$)=0\mathrm{.}1,$ P$($miss $->$

attend$)=0\mathrm{.}5,$ P$($miss $->$ miss$)=0\mathrm{.}5$