Bayes Theorem
Reuben
2024-05-24
GUIDED PRACTICE 3.43 Q) Jose visits campus every Thursday evening. However, some days the parking garage is full, often due to college events. There are academic events on 35% of evenings, sporting events on 20% of evenings, and no events on 45% of evenings. When there is an academic event, the garage fills up about 25% of the time, and it fills up 70% of evenings with sporting events. On evenings when there are no events, it only fills up about 5% of the time. If Jose comes to campus and finds the garage full, what is the probability that there is a sporting event? Use a tree diagram to solve this problem.
library(Rgraphviz)## Loading required package: graph## Loading required package: BiocGenerics##
## Attaching package: 'BiocGenerics'## The following objects are masked from 'package:stats':
##
## IQR, mad, sd, var, xtabs## The following objects are masked from 'package:base':
##
## anyDuplicated, aperm, append, as.data.frame, basename, cbind,
## colnames, dirname, do.call, duplicated, eval, evalq, Filter, Find,
## get, grep, grepl, intersect, is.unsorted, lapply, Map, mapply,
## match, mget, order, paste, pmax, pmax.int, pmin, pmin.int,
## Position, rank, rbind, Reduce, rownames, sapply, setdiff, table,
## tapply, union, unique, unsplit, which.max, which.min## Loading required package: gridlibrary(BiocManager)#Sport event (A1) :
A1 = 20
#Academic event (A2) :
A2 = 35
#No events (A3) :
A3 = 45
#Find Probability of A1, A2, A3
#Probability of A1 :
PA1 = A1/100
print(PA1)## [1] 0.2## [1] 0.2
nota1 = 1-PA1
#Probability of A2 :
PA2 = A2/100
print(PA2)## [1] 0.35## [1] 0.35
nota2 = 1-PA2
#Probability of A3 :
PA3 = A3/100
print(PA3)## [1] 0.45## [1] 0.45
nota3 = 1-PA3
#Find the probability of garage full for A1, A2, A3 :
b1 = 70
b2 = 25
b3 = 5
#Probability of (B/A1) :
B1A1 = b1/100
print(B1A1)## [1] 0.7## [1] 0.7
#Probability of (B/A2) :
B2A2 = b2/100
print(B2A2)## [1] 0.25## [1] 0.25
#Probability of (B/A3) :
B3A3 = b3/100
print(B3A3)## [1] 0.05## [1] 0.05
#Not functions of B1A1, B2A2, B3A3 :
notb1a1 = 1-B1A1
notb2a2 = 1-B2A2
notb3a3 = 1-B3A3
#Find the probability with Bayes theorem
#The formula is P(A1/B1) = P(B1A1)*P(A1) / P(B1A1)*P(A1)+P(B2A2)*P(A2)+P(B3A3)*P(A3)
#After the apply all the values in the formula :
aAndb = B1A1*PA1
aAndnotb = notb1a1*PA1
a2Andb2 = B2A2*PA2
a2Andnotb2 = notb2a2*PA2
a3Andb3 = B3A3*PA3
a3Andnotb3 = notb3a3*PA3
Top = (B1A1) * (PA1)
print(Top)## [1] 0.14## [1] 0.14
Bot1 = B2A2 * PA2
print(Bot1)## [1] 0.0875## [1] 0.0875
Bot2 = B3A3 * PA3
print(Bot2)## [1] 0.0225## [1] 0.0225
plus = Top + Bot1 + Bot2
print(plus)## [1] 0.25## [1] 0.25
PA1B1 = Top / plus
print(PA1B1)## [1] 0.56## [1] 0.56
# (0.7)(0.2) / (0.7)(0.2)(0.25)(0.35)+(0.05)(0.45) = 0.140 / .25 = 0.56.
# The probability of sporting event is 0.56.
# Creating Tree Diagram :
node1 = "P"
node2 = "A1"
node3 = "A2"
node4 = "A3"
node5 = "A1&B"
node6 = "A1&B'"
node7 = "A2&B"
node8 = "A2&B'"
node9 = "A3&B"
node10 = "A3&B'"
nodeNames = c(node1,node2,node3,node4, node5,node6, node7, node8, node9, node10)
rEG <- new("graphNEL", nodes=nodeNames, edgemode="directed")
# Draw the "lines" or "branches" of the probability Tree
rEG <- addEdge(nodeNames[1], nodeNames[2], rEG, 1)
rEG <- addEdge(nodeNames[1], nodeNames[3], rEG, 1)
rEG <- addEdge(nodeNames[1], nodeNames[4], rEG, 1)
rEG <- addEdge(nodeNames[2], nodeNames[5], rEG, 1)
rEG <- addEdge(nodeNames[2], nodeNames[6], rEG, 1)
rEG <- addEdge(nodeNames[3], nodeNames[7], rEG, 1)
rEG <- addEdge(nodeNames[3], nodeNames[8], rEG, 1)
rEG <- addEdge(nodeNames[4], nodeNames[9], rEG, 1)
rEG <- addEdge(nodeNames[4], nodeNames[10], rEG, 10)
eAttrs <- list()
q<-edgeNames(rEG)
# Add the probability values to the the branch lines
eAttrs$label <- c(toString(A1),toString(A2),
toString(A3), toString(B1A1),
toString(notb1a1), toString(B2A2), toString(notb2a2), toString(B3A3), toString(notb3a3))
names(eAttrs$label) <- c(q[1],q[2], q[3], q[4], q[5], q[6], q[7], q[8], q[9])
edgeAttrs<-eAttrs
# Color of tree
attributes<-list(node=list(label="foo", fillcolor="lightgreen", fontsize="15"),
edge=list(color="red"),graph=list(rankdir="LR"))
#Plot
plot(rEG, edgeAttrs=eAttrs, attrs=attributes)
nodes(rEG)## [1] "P" "A1" "A2" "A3" "A1&B" "A1&B'" "A2&B" "A2&B'" "A3&B"
## [10] "A3&B'"## [1] "P" "A1" "A2" "A3" "A1&B" "A1&B'" "A2&B" "A2&B'" "A3&B"
## [10] "A3&B'"
edges(rEG)## $P
## [1] "A1" "A2" "A3"
##
## $A1
## [1] "A1&B" "A1&B'"
##
## $A2
## [1] "A2&B" "A2&B'"
##
## $A3
## [1] "A3&B" "A3&B'"
##
## $`A1&B`
## character(0)
##
## $`A1&B'`
## character(0)
##
## $`A2&B`
## character(0)
##
## $`A2&B'`
## character(0)
##
## $`A3&B`
## character(0)
##
## $`A3&B'`
## character(0)## $P
## [1] "A1" "A2" "A3"
##
## $A1
## [1] "A1&B" "A1&B'"
##
## $A2
## [1] "A2&B" "A2&B'"
##
## $A3
## [1] "A3&B" "A3&B'"
##
## $`A1&B`
## character(0)
##
## $`A1&B'`
## character(0)
##
## $`A2&B`
## character(0)
##
## $`A2&B'`
## character(0)
##
## $`A3&B`
## character(0)
##
## $`A3&B'`
## character(0)
text(570,390,aAndb, cex=.8)
text(570,320,aAndnotb,cex=.8)
text(570,245,a2Andb2,cex=.8)
text(570,170,a2Andnotb2,cex=.8)
text(570,95,a3Andb3,cex=.8)
text(570,30,a3Andnotb3,cex=.8)
text(350,385,"(B1 | A1)",cex=.8)
text(350,255,"(B2 | A2)",cex=.8)
text(350,120,"(B3 | A3)",cex=.8)
#Write a table in the lower left of the probablites of A and B
text(50,50,paste("P(A):",PA1),cex=.9, col="darkgreen")
text(50,20,paste("P(A'):",nota1),cex=.9, col="darkgreen")
text(110,50,paste("P(B):",round(PA2,digits=2)),cex=.9, col="darkgreen")
text(110,20,paste("P(B'):",round(nota2, 2)),cex=.9, col="darkgreen")
text(175,50,paste("P(C):",round(PA3,digits=2)),cex=.9, col="darkgreen")
text(175,20,paste("P(C'):",round(nota3, 2)),cex=.9, col="darkgreen")
text(80,420,paste("P(A|B): ",round(PA1B1,digits=2)),cex=.9,col="blue")