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$)=0\mathrm{.}5$

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