# 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