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# В следващите задачи намерете приближение на търсените вероятности чрез подходящи симулации
ex1 <- function() {
balls <- sample(seq(1, 8, 1), 2, replace=T)
balls[1] == balls[2]
}
sum(replicate(10000, ex1())) / 10000
ex2 <- function() {
socks <- sample(rep(seq(1, 3, 1), times=2), 2, replace=F)
socks[1] == socks[2]
}
sum(replicate(10000, ex2())) / 10000
ex3 <- function() {
tried_keys <- sample(c(1, 0, 0, 0), 4, replace=F)
tried_keys[4]
}
sum(replicate(100000, ex3())) / 100000
ex4 <- function() {
questions <- sample(append(rep(1, times=17), rep(0, times=3)), 2, replace=F)
sum(questions) == 1
}
sum(replicate(100000, ex4())) / 100000
ex5 <- function() {
birthdays <- sample(seq(1, 365, 1), 25, replace=T)
sum(duplicated(birthdays)) > 0
}
sum(replicate(100000, ex5())) / 100000
ex6 <- function() {
presents <- sample(seq(1, 20, 1), 20, replace=F)
0 %in% (presents - seq(1, 20, 1))
}
sum(replicate(100000, ex6())) / 100000
ex7 <- function() {
ants <- c(sample(c(2, 3), 1), sample(c(1, 3), 1), sample(c(2, 3), 1))
sum(ants) == 6
}
sum(replicate(100000, ex7())) / 100000
ex8 <- function(cmp) {
eggs <- c(rep(0, times=6), rep(1, times=2))
samp <- sample(eggs, 8, replace=F)
cmp(samp[seq(1, 7, 2)], samp[seq(2, 8, 2)])
}
sum(replicate(100000, ex8(function(first, second) { sum(first) == 2 | sum(second) == 2 }))) / 100000
sum(replicate(100000, ex8(function(first, second) { sum(first) == 1 }))) / 100000
sum(replicate(100000, ex8(function(first, second) { sum(first) == 2 }))) / 100000
sum(replicate(100000, ex8(function(first, second) { sum(second) == 1 }))) / 100000
ex9 <- function() {
answers <- sample(c(0, 0, 0, 1), 10, replace=T)
sum(answers) >= 5
}
sum(replicate(100000, ex9())) / 100000
ex10 <- function() {
ontime <- sample(c(0, 1), 143, replace=T, prob=c(0.08, 0.92))
sum(ontime) <= 138
}
sum(replicate(100000, ex10())) / 100000
ex11 <- function() {
if (sample(seq(1, 6, 1), 1) == 6) {
sample(c(1, 1, 0, 0), 1) == 1
}
else {
sample(c(1, 0, 0, 0, 0), 1) == 1
}
}
sum(replicate(100000, ex11())) / 100000
ex12 <- function() {
coin <- sample(c(11, 11, 22, 12, 12), 1)
if (coin == 11) { T }
else if (coin == 22) { F }
else { sample(c(T, F), 1) }
}
sum(replicate(100000, ex12())) / 100000
# В следващите задачи намерете вероятностите като използвате вградените функции, свързани с дискретни разпределения.
# P(X = x) := d...(x, ...)
# P(X <= x) := p...(x, ...)
# ex18
# Биномно : брой успехи от n опита
dbinom(2, 10, 1/6)
pbinom(2, 10, 1/6)
1 - pbinom(2, 10, 1/6)
pbinom(8, 10, 1/6) - pbinom(3, 10, 1/6)
# ex19
# Геометрично : брой **неуспехи** до първи успех
pgeom(9, 1/6)
1 - pgeom(4, 1/6)
# ex20
# Отрицателно биномно : брой **неуспехи** до r-ти успех
1 - pnbinom(20-3, 3, 1/6)
# ex21
# Хипергеометрично : от маркирани и немаркирани в извадка, броя маркирани
1 - dhyper(2, 5, 3, 2)
# ex22
# Поасоново : средно толкова събитие на единица време, колко събития за мн време
ppois(2, 5 * 300 / 500)
ppois(3, 5 * 300 / 500) - ppois(0, 5 * 300 / 500)
# ex23
1 - dbinom(4, 10, 1/4)
# ex24
pbinom(138, 143, 0.92)
dbinom(137, 143, 0.92)
# ex25
pgeom(10-1, 0.03)
1 - dbinom(1, 50, 0.03)
# ex26
1 - phyper(1, 3, 97, 50)
# ex27
1 - phyper(1, 90, 2910, 50)
# ex28
1 - ppois(4, 1/10)
1 - pnbinom(39, 10, 1/10)
# ***
# ex32
runif(500, min=2, max=3)
# ex38
A.p <- 1 - pnorm(250, 252, 3)
B.p <- pbinom(2, 5, A.p)
# ex39
# ***
# ex67
# успех, общо--| |--хипотеза---|
prop.test(32, 58, 0.51, "greater")
# prop.test проверява дали вероятност е валидна
# ex68
# нормално разпределение
t.test(c(3.1, 3.0, 3.7, 2.6, 4.2, 3.8, 3.6, 2.7, 3.8, 4.4), mu=4, alternative="less")
# t.test проверява средните стойности между две м-ва данни
# paired=T ако са зависими
# ***
# ex88
chisq.test(c(28, 36, 36, 30, 27, 23), p=rep(1/6, times=6))
# chisq.test проверява измервания с мат модел
# Минали изпити
# ex1
ex1 <- function() {
throws <- sample(seq(1, 6, 1), 4, replace=T)
sum(throws[c(1,2)]) > sum(throws[c(3,4)])
}
sum(replicate(100000, ex1())) / 100000
# ex2
prop.test(15, 500, p=0.035, alternative="less", correct=T)
# alternative is "less", so first one
# 0.3132 > 0.05 => null hypothesis is correct => alternative is false
# ex3
t.test(a1, a2, alternative="greater", paired=T)
# paired t-test е, оценката по алгебра 1 и алгебра 2 са зависими
# ex4
# оценено регресионно уравнение
# grade = beta_0 + beta_1 * study.hours + epsilon
# grade = 1.01617 + 0.15806 * study.hours
# за всеки час учене, оценката се вдига с 0.15806
# 2.9e-12 < 0.05 => null hypothesis is incorrect => alternative is true
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