Question 1:
In an office, there are 10 sales person,four women and six men. Three women and two men will be chosen at random. Use simulations in R to approximate the probability.
nloop=100000
sample1 = c(rep("W",4),rep("M",6))
sample1## [1] "W" "W" "W" "W" "M" "M" "M" "M" "M" "M"count = 0
for (iloop in 1:nloop){
MW = sample(sample1,5, replace=F)
if (sum(length(which("M"==MW)))==2){count =count+1}
}
count/nloop## [1] 0.23947#Mathematical Solution
Com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
(Com(4,3) * Com(6,2))/ Com(10,5)## [1] 0.2380952Question 2:
There are 7 balls in a urn, they are identical except color, 3 blue and 4 yellow. What is the probability of choosing 2 balls in a different color without replacement ? Write R code to approximate the probability.
nloop=100000
bag = c(rep(1,3),rep(0,4))
#Let's assign the colors to numbers: Blue=1 and Yellow=0
count=0
for (iloop in 1:nloop){
tot = sample(bag,2, replace=F)
if (sum(tot)==1 ){count =count+1}
}
count/nloop## [1] 0.57081#Mathematical Solution
Com(2,1)*(3/7)*(4/6)## [1] 0.5714286Question 3:
There are 10 balls in each urn. 5 blue and 5 yellow ball in the 1st urn, and 4 blue, and 6 yellow in the 2nd urn. Sandy picks 1 ball from each urn at random. What is the probability of choosing the balls in the same color? Write R code to approximate the probability.
# Let's code Blue= 1, Yellow= 0
nloop=100000
sam1 = c(rep(1,5),rep(0,5))
sam2 = c(rep(1,4),rep(0,6))
count=0
for (iloop in 1:nloop){
urn1 = sample(sam1,1, replace=F)
urn2 = sample(sam2,1, replace=F)
urn = urn1 + urn2
if (sum(urn)==2 | sum(urn==0)){count =count+1}
}
count/nloop## [1] 0.50089#Mathematical Solution
(5/10)*(4/10)+(5/10)*(6/10)## [1] 0.5Question 4: Suppose you roll a fair six-sided die six times. Write R code to approximate the probability that each number comes up exactly one time.
nloop=100000
sam1 = rep(1:6,1)
count=0
for (iloop in 1:nloop){
roll = sample(sam1,6, replace=T)
sroll = sort(roll)
if (all(sroll == c(1,2,3,4,5,6)) ){count =count+1}
}
count/nloop## [1] 0.01489#Mathematical Solution
(6/6)*(5/6)*(4/6)*(3/6)*(2/6)*(1/6)## [1] 0.0154321Question 5: What is the probability of a large straight when five dice are rolled? This is defined (in the game Yahtzee) as the numbers 1,2,3,4,5 or 2,3,4,5,6. Write R code to approximate this probability.
nloop=100000
sam1 = rep(1:6,1)
count=0
for (iloop in 1:nloop){
roll = sample(sam1,5, replace=T)
sroll = sort(roll)
if (all(sroll == c(1,2,3,4,5)) | all(sroll == c(2,3,4,5,6))){count =count+1}
}
count/nloop## [1] 0.03032#Mathematical Solution
2*(5/6)*(4/6)*(3/6)*(2/6)*(1/6)## [1] 0.0308642Question 6: For a recent year, 51% of the families in the United States had no children under the age of 18; 20% had one child; 19% had two children; 7% had three children; and 3% had four or more children.
If three family is selected at random, find the probability that total number of children these families have less then 7. Write R code to determine the probability. Use simulations in R to approximate probability.
nloop = 100000
count = 0
for (iloop in 1:nloop){
child = sample(0:4,3,replace = T, prob = c(.51,.2,.19,.07,.03))
if (sum(child)<7){count=count+1}
}
count/nloop## [1] 0.96049#Mathematical Solution
# We need to find all different possibilities and add them up. (000,001,002,...)Question 7: Four children from a neighborhood in Smalltown attend a holiday party. There are 40 children total at the party, and 20 prizes are to be distributed at random to the children. That is, each prize is equally likely to go to each of the 40 children, regardless of whether a child already has prizes.
#a.
nloop =100000
count=0
for(iloop in 1:nloop){
prize=sample(1:40,20,replace=T)
s_prize = sort(prize)
if (all(s_prize[1:4] == c(1,2,3,4)) & s_prize[5]!=4){count= count+1}
}
count/nloop## [1] 0.00816#Mathematical Solution
(20*19*18*17*(36)^16)/40^20## [1] 0.008416765nloop =100000
count=0
for(iloop in 1:nloop){
price=sample(1:40,20,replace=T)
if (all(c(1,2,3,4) %in% price)){count= count+1}
}
count/nloop## [1] 0.02053#Mathematical Solution
(20*19*18*17*(40)^16)/40^20## [1] 0.04542188#c. UNIQUE and LENGTH
nloop =100000
count=0
for(iloop in 1:nloop){
prize=sample(1:40,20,replace=T)
u_prize =length(unique(prize))
if (u_prize== 20){count= count+1}
}
count/nloop## [1] 0.00288#Mathematical Solution
(factorial(20)*Com(40,20))/40^20## [1] 0.003050146Question 8: Two factories supply light bulbs to the market. Bulbs from factory X work for over 6000 hours in 98% of cases, whereas bulbs from factory Y work for over 6000 hours in 93% of cases. It is known that
factory X supplies 65% of the total bulbs available in the market. Given that a light bulb works for more than 6000 hours, what is the probability that it was supplied by factory Y ? Use simulations in R to approximate probability.
nloop =100000
countx =0
county =0
countxy =0
for (iloop in 1:nloop){
xy= sample(c("x","y"),1, prob=c(0.65,0.35))
if (xy=="x"){
x = sample(0:1,1, prob =c(0.02,0.98))
if(x=="1"){countx =countx+1}
}
if (xy=="y"){
y = sample(0:1,1, prob =c(0.07,0.93))
if(y=="1"){county = county+1}
}
}
county/(countx+county)## [1] 0.3392811#Mathematical Solution
(93*35)/(98*65+93*35)## [1] 0.3381818Question 9:
A drawer contains 10 pairs of socks, where each pair is a different color. Sam draws four socks at random from the drawer (without replacement). Write R code to determine the probability that there is no pair among the four socks.
nloop =100000
count =0
socks = c(rep(1:10,each=2))
for (iloop in 1:nloop){
draw_soc = sample(socks,4, replace=F)
sock = unique(draw_soc)
if(length(sock)==4){count= count+1}
}
count/nloop## [1] 0.69027#Mathmatical Solution:
(20/20)*(18/19)*(16/18)*(14/17)## [1] 0.6934985Question 10:
There are 4 girl and 5 boy in the classroom. What is the probability of having girls stand next to each other in the line? Use simulations in R to approximate probability.
nloop =100000
count =0
tot = c(1:9)
for (iloop in (1:nloop-3)){
samp = sample(tot,9, replace=F)
for (i in 1:9){
if(all(samp[i] %in% c(1:4) &
samp[i+1] %in% c(1:4) &
samp[i+2] %in% c(1:4) &
samp[i+3] %in% c(1:4))) {count =count+1}
}
}
count/nloop## [1] 0.04708#Mathematical Solution
factorial(6)*factorial(4)/factorial(9)## [1] 0.04761905Question 11:
In a classroom, there are total 30 students and 8 of them have glasses, and half of the students who wear and not wear glasses habe brown eyes. What is the probability of choosing a student who has a broen eye or wear glasses?
nloop=100000
count =0
class = c(rep(1,19),rep(0,11))
for (iloop in 1:nloop){
bg= sample(class,1, replace=F)
if(bg==1){count=count+1}
}
count/nloop## [1] 0.63619#Mathematical solution
19/30## [1] 0.6333333Question 12:
The probablity of a hunter to hit the targer is 3/4. If he makes 4 shoot, what is the probability of hitting the target only on shoot 2 and shoot 4?
nloop =100000
count =0
for (iloop in 1:nloop){
hit1 = sample(0:1,1,replace= F,prob=c(.25,.75))
hit2 = sample(0:1,1,replace= F,prob=c(.25,.75))
hit3 = sample(0:1,1,replace= F,prob=c(.25,.75))
hit4 = sample(0:1,1,replace= F,prob=c(.25,.75))
if((sum(hit1+hit3)==0) & (sum(hit2+hit4)==2)){count=count+1}
}
count/nloop## [1] 0.03554#Mathematical Solution
(1/4)*(3/4)*(1/4)*(3/4)## [1] 0.03515625Question 13:
Two evenly matched baseball teams play each other for a series of seven games. Estimate the probability that Team I will win the series by winning at least four games from Team 2. Use simulations in R to approximate wining probability.
nloop = 100000
count =0
for (iloop in 1: nloop){
win = sample(0:1,7,replace=T)
if (sum(win)>=4){count =count+1}
}
count/nloop## [1] 0.50152#Mathematical Solution
com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
com(7,4)*(1/2)^7+com(7,5)*(1/2)^7+com(7,6)*(1/2)^7+com(7,7)*(1/2)^7## [1] 0.5Question 14:
Five team joins a soccer tournamnet. The wining posibilities as follows: Team A & Team B =60%, Team A & Team C =40%, Team A & Team D =80%, Team A & Team E =30%, Team B & Team C =30%, Team C & Team D =70%, Team C & Team E =40%, Team C & Team D =50%, Team C & Team E =80%, Team D & Team E =90%,
Use simulations in R to approximate the winning posibilities for each team.
nloop =100
countA =0; countB =0; countC =0; countD =0; countE =0
for (i in 1:nloop){
gameAB= sample(c(1,2),1, prob=c(0.6,0.4))
gameAC= sample(c(1,3),1, prob=c(0.4,0.6))
gameAD= sample(c(1,4),1, prob=c(0.8,0.2))
gameAE= sample(c(1,5),1, prob=c(0.3,0.7))
gameBC= sample(c(2,3),1, prob=c(0.3,0.7))
gameBD= sample(c(2,4),1, prob=c(0.7,0.3))
gameBE= sample(c(2,5),1, prob=c(0.4,0.6))
gameCD= sample(c(3,4),1, prob=c(0.5,0.5))
gameCE= sample(c(3,5),1, prob=c(0.8,0.2))
gameDE= sample(c(4,5),1, prob=c(0.9,0.1))
games =c(gameAB,gameAC,gameAD,gameAE,gameBC,gameBD,gameBE,gameCD,gameCE,gameDE)
countA = countA + length(which(1==games))
countB = countB + length(which(2==games))
countC = countC + length(which(3==games))
countD = countD + length(which(4==games))
countE = countE + length(which(5==games))
}
print(paste("Team A :", countA,"// Team B :",countB, "// Team C :",countC,"// Team D :",countD,"// Team E :",countE))## [1] "Team A : 223 // Team B : 193 // Team C : 229 // Team D : 175 // Team E : 180"Question 15:
The experiment is to roll four fair, six-sided dice. Given that all dice show numbers that are three or greater, what is the probability that all the dice show either odd numbers or even numbers(3,5,7,5 or 4,6,4,6) ? Use simulations in R to compute this conditional probability.
nloop=100000; countA=0; countAB=0
for(iloop in 1:nloop){
dice=sample(1:6,4,replace=TRUE)
eventA = all(dice>=3)
eventB = (all(dice%%2==1) | all(dice%%2==0))
if(eventA){countA=countA+1}
if(eventA&eventB){countAB=countAB+1}
}
countAB/countA## [1] 0.1218972# Mathematical Solution
# Choosing odd numbers =2^4
# Chosing even numbers =2^4
# All possibilites that are greater or equal to 3 = (4)^4 =256
(2^4+2^4)/4^4## [1] 0.125Question 16:
A quiz consists of ten multiple choice questions with choices (a) and (b)for each. If an unprepared student marks answers at random, what are the probabilities of the following events:
Use simulations in R to approximate probabilities.
nloop =100000
count =0
for (iloop in 1:nloop){
quiz = sample(0:1,10,replace = T)
#Let's choose 1's are correct and 0's are wrong answer
if (sum(quiz)==1){count= count+1}
}
count/nloop## [1] 0.00953#Mathematical Solution
Com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
(Com(10,1) )/ 2^10## [1] 0.009765625nloop =100000
count =0
for (iloop in 1:nloop){
quiz = sample(0:1,10,replace = T)
#Let's choose 1's are correct and 0's are wrong answer
if (sum(quiz)>=7){count= count+1}
}
count/nloop## [1] 0.17225#Mathematical Solution
Com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
(Com(10,7) + Com(10,8) + Com(10,9) + Com(10,10))/ 2^10## [1] 0.171875Question 17:
A quiz consists of five multiple choice questions with choices (a), (b) and (c)for each. If an unprepared student marks answers at random, what are the probabilities of the following events:
Use simulations in R to approximate probabilities.
nloop =100000
count =0
for (iloop in 1:nloop){
quiz = sample(c("a",'b',"c"),5,replace = T)
#Let's choose a's are correct and b's and c's are wrong answer
if (quiz[1]=="a"){count= count+1}
}
count/nloop## [1] 0.33286#Mathematical Solution
Com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
(Com(5,1)*(1/3)*(2/3)^4)## [1] 0.3292181nloop =100000
count =0
for (iloop in 1:nloop){
quiz = sample(c("a",'b',"c"),5,replace = T)
#Let's choose a's are correct and b's and c's are wrong answer
if ((sum(sapply(quiz=="a",unique)))>=3){count= count+1}
}
count/nloop## [1] 0.2104#Mathematical Solution
Com = function(n, x) {
factorial(n) / factorial(n-x) / factorial(x)
}
(Com(5,3)*(1/3)^3*(2/3)^2 + Com(5,4)*(1/3)^4*(2/3) + Com(5,5)*(1/3)^5)## [1] 0.2098765Question 18:
Simulate Birthday paradox with R.
The paradox of birthdays is a mathematical problem put forward by Von Mises, who looks for the value N in the problem: In a group of N people there is 50% chance that at least 2 people in the group share the same birthday (day + month). The answer is N = 23, which is quite counterintuitive, hence the paradox.
# Birthday Function
birthday <- function(n){
f=1
for(i in 1:n){
f <- f* (365-i+1)/365
}
print(paste("Probability of",i,"people :",1-f))
}
# Birthday Paradox Simulation
nloop=50
birthstat=1:nloop
for (i in 1:nloop){
birthstat[i]=birthday(i)
}## [1] "Probability of 1 people : 0"
## [1] "Probability of 2 people : 0.00273972602739725"
## [1] "Probability of 3 people : 0.00820416588478146"
## [1] "Probability of 4 people : 0.0163559124665503"
## [1] "Probability of 5 people : 0.0271355736997937"
## [1] "Probability of 6 people : 0.0404624836491116"
## [1] "Probability of 7 people : 0.0562357030959756"
## [1] "Probability of 8 people : 0.0743352923516692"
## [1] "Probability of 9 people : 0.094623833889167"
## [1] "Probability of 10 people : 0.116948177711078"
## [1] "Probability of 11 people : 0.141141378321733"
## [1] "Probability of 12 people : 0.167024788838065"
## [1] "Probability of 13 people : 0.19441027523243"
## [1] "Probability of 14 people : 0.223102512004973"
## [1] "Probability of 15 people : 0.252901319763687"
## [1] "Probability of 16 people : 0.28360400525285"
## [1] "Probability of 17 people : 0.315007665296561"
## [1] "Probability of 18 people : 0.346911417871789"
## [1] "Probability of 19 people : 0.379118526031537"
## [1] "Probability of 20 people : 0.41143838358058"
## [1] "Probability of 21 people : 0.443688335165206"
## [1] "Probability of 22 people : 0.47569530766255"
## [1] "Probability of 23 people : 0.507297234323986"
## [1] "Probability of 24 people : 0.538344257914529"
## [1] "Probability of 25 people : 0.568699703969464"
## [1] "Probability of 26 people : 0.598240820135939"
## [1] "Probability of 27 people : 0.626859282263242"
## [1] "Probability of 28 people : 0.654461472342399"
## [1] "Probability of 29 people : 0.680968537477777"
## [1] "Probability of 30 people : 0.706316242719269"
## [1] "Probability of 31 people : 0.730454633728644"
## [1] "Probability of 32 people : 0.753347527850321"
## [1] "Probability of 33 people : 0.774971854175772"
## [1] "Probability of 34 people : 0.795316864620154"
## [1] "Probability of 35 people : 0.814383238874715"
## [1] "Probability of 36 people : 0.83218210637988"
## [1] "Probability of 37 people : 0.848734008216385"
## [1] "Probability of 38 people : 0.864067821082121"
## [1] "Probability of 39 people : 0.878219664366722"
## [1] "Probability of 40 people : 0.891231809817949"
## [1] "Probability of 41 people : 0.903151611481735"
## [1] "Probability of 42 people : 0.914030471561869"
## [1] "Probability of 43 people : 0.92392285565612"
## [1] "Probability of 44 people : 0.932885368551426"
## [1] "Probability of 45 people : 0.940975899465775"
## [1] "Probability of 46 people : 0.948252843367255"
## [1] "Probability of 47 people : 0.954774402833299"
## [1] "Probability of 48 people : 0.960597972879422"
## [1] "Probability of 49 people : 0.965779609322676"
## [1] "Probability of 50 people : 0.970373579577988"# x=1:50
# plot(x,birthstat, main= "Birthday Paradox", xlab = "Number of people", ylab="Probability")
# abline(h=0.5, xlim =c(0,50),col="red")