Week 5 Discussion
2023 10 2
Haiding Luo
Part I
I. Please explain Bayes Theorem in your own words, and give an example. Less than 10 sentences. Also, write out the formula. Pick up on how to to type equations in R Markdown using Latex terminology
Bayes’ Theorem is a mathematical formula based on conditional
probability, used to estimate the probability of one event occurring
given the occurrence of another related event. The formula for Bayes’
Theorem is:
\[ P(A \mid B) = \frac{P(A)*P(B \mid A)}{P(B)} \]
where P(A) and P(B) represent the probabilities of events A and B occurring, respectively. P(B|A) represents the conditional probability of event B occurring given that event A has occurred, and P(A|B) represents the conditional probability of event A occurring given that event B has occurred.
Example:
There are two buckets. Bucket 1 contains 40 balls, including 30 white balls and 10 black balls. Bucket 2 also contains 40 balls, with 20 white balls and 20 black balls.
Event A: Drawing 1 ball from bucket 1. Event B: Drawing a white ball. Event C: Drawing 1 ball from bucket 2.
If I grab a white ball, what is the probability P(A|B) that the white ball came from bucket 1?
\[ P(B \mid A)=30/(30+10) = 75\% \]
\[ P(B \mid C) = 20/(20+20)= 50\% \]
\[ P(B) = P(B \mid A) P(A)+P(B \mid C)P(C)=75\%*50\% + 50\%* 50\% = 62.5\% \]
\[ \frac{P(B \mid A)}{P(B)}= 75\% / 62.5 \% = 1.2 \]
\[ \frac{P(A)*P(B \mid A)}{P(B)} = \frac{1}{2}* 1.2 = \frac{3}{5} \]
Part II
options(repos = c(CRAN = "https://cran.rstudio.com/"))
install.packages("BiocManager")## 将程序包安装入'C:/Users/pokem/AppData/Local/R/win-library/4.3'
## (因为'lib'没有被指定)## 程序包'BiocManager'打开成功，MD5和检查也通过
##
## 下载的二进制程序包在
## C:\Users\pokem\AppData\Local\Temp\RtmpohcMCT\downloaded_packages里library(BiocManager)
BiocManager::install("Rgraphviz")## 'getOption("repos")' replaces Bioconductor standard repositories, see
## 'help("repositories", package = "BiocManager")' for details.
## Replacement repositories:
## CRAN: https://cran.rstudio.com/## Bioconductor version 3.17 (BiocManager 1.30.22), R 4.3.1 (2023-06-16 ucrt)## Warning: package(s) not installed when version(s) same as or greater than current; use
## `force = TRUE` to re-install: 'Rgraphviz'## Installation paths not writeable, unable to update packages
## path: C:/Program Files/R/R-4.3.1/library
## packages:
## foreign, KernSmooth, lattice, Matrix, mgcv, nlme, spatial, survival## Old packages: 'curl', 'evaluate', 'Hmisc', 'knitr', 'lubridate', 'markdown',
## 'openssl', 'plyr', 'psych', 'renv', 'rmarkdown', 'tinytex', 'withr'library(Rgraphviz)## 载入需要的程辑包：graph## 载入需要的程辑包：BiocGenerics##
## 载入程辑包：'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, sort,
## table, tapply, union, unique, unsplit, which.max, which.min## 载入需要的程辑包：gridGuided Practice 3.43: 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.
Let:
A be the event of an academic event.
S be the event of a sporting event.
N be the event of no event.
F be the event of the garage being full.
P(A) = 0.35
P(S) = 0.20 P(N) = 0.45
P(F|A) = 0.25
P(F|S) = 0.70
P(F|N) = 0.05
\[ P(S \mid F) =\frac{P(S)*P(F\mid S)}{P(A)*P(F\mid A)+P(S)*P(F\mid S)+P(N)*P(F\mid N)} \]
P_A <- 0.35
P_S <- 0.20
P_N <- 0.45
P_F_A <- 0.25
P_F_S <- 0.70
P_F_N <- 0.05
Prob <- (P_S * P_F_S) / (P_A * P_F_A + P_S * P_F_S + P_N * P_F_N)
Prob## [1] 0.56a1 <- .35
a2 <- .20
a3 <- .45
fa1 <- .25
fa2 <- .7
fa3 <- .05
notfa1 <- 1 - fa1
notfa2 <- 1 - fa2
notfa3 <- 1 - fa3
aANDfa1 <- a1 * fa1
bANDfa2 <- a2 * fa2
cANDfa3 <- a3 * fa3
node1 <- "P"
node2 <- "a1"
node3 <- "a2"
node4 <- "a3"
node5 <- "a1ANDfa1"
node6 <- "notfa1"
node7 <- "a2ANDfa2"
node8 <- "notfa2"
node9 <- "a3ANDfa3"
node10 <- "notfa3"
nodeNames <- c(node1, node2, node3, node4, node5, node6, node7, node8, node9, node10)
rEG <- new("graphNEL",
nodes = nodeNames,
edgemode="directed"
)
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)
eAttrs$label <- c(toString(a1), toString(a2),
toString(a3), toString(fa1),
toString(notfa1), toString(fa2),
toString(notfa2), toString(fa3),
toString(notfa3)
)
names(eAttrs$label) <- c( q[1], q[2], q[3], q[4], q[5], q[6], q[7], q[8], q[9])
edgeAttrs <- eAttrs
attributes <- list(node = list(label = "foo",
fillcolor = "green",
fontsize = "15"
),
edge = list(color = "red"),
graph = list(rankdir = "LR")
)
plot (rEG, edgeAttrs = eAttrs,attrs=attributes)
text(578,410, aANDfa1, cex = .8)
text(570,320,notfa1,cex=.8)
text(578,230, bANDfa2, cex = .8)
text(570,170,notfa2,cex=.8)
text(578,95, cANDfa3, cex = .8)
text(570,30,notfa3,cex=.8)
text(160,55, paste('P(B):', a2), cex = 1.1)
text(160,35, paste('P(F):', aANDfa1+bANDfa2+cANDfa3), cex = 1.1)
text(160,15, paste('P(B|F):', fa2*a2/(aANDfa1+bANDfa2+cANDfa3)), cex = 1.1)