HW 5

Prove that B and C are proper probability distributions

#randomly select 20,000 numbers between 0 and 1
B <- runif(10000, min = 0, max = 1)
head(B)
## [1] 0.9661127 0.2874410 0.5546079 0.5063111 0.8729753 0.2976092
C <- runif(10000, min = 0, max = 1)
head(C)
## [1] 0.02010106 0.54572742 0.43033592 0.89223379 0.33233312 0.32323655
head(B+C)
## [1] 0.9862138 0.8331684 0.9849438 1.3985449 1.2053084 0.6208458

All values of B and C are between 0 and 1 and sum of the probabilites of each value in B and C will equal 1.

B+C < 1/2

a <- sum((B+C) < .5)/10000
print(paste("The probability that B+C will be less than 1/2 is",a))
## [1] "The probability that B+C will be less than 1/2 is 0.1259"

BC < 1/2

b <- (sum((B*C) < .5)/10000)
print(paste("The probability that B*C will be less than 1/2 is",b))
## [1] "The probability that B*C will be less than 1/2 is 0.8437"

|B-C| < 1/2

c <- sum(abs((B-C)) < .5)/10000
print(paste("The probability that |B-C| will be less than 1/2 is",c))
## [1] "The probability that |B-C| will be less than 1/2 is 0.7489"

max{B,C} < 1/2

x <- 0
for(i in 1:10000){
  if(max(c(B[i],C[i])) < 0.5){
    x = x+1
  }
}
d <- x/10000
print(paste("The probability that max{B,C} will be less than 1/2 is",d))
## [1] "The probability that max{B,C} will be less than 1/2 is 0.2532"

min{B,C} < 1/2

x <- 0
for(i in 1:10000){
  if(min(c(B[i],C[i])) < 0.5){
    x = x+1
  }
}
e <- x/10000
print(paste("The probability that min{B,C} will be less than 1/2 is",e))
## [1] "The probability that min{B,C} will be less than 1/2 is 0.7532"