HW4

mat <- matrix(1:6)
# create a dice
n_simulations <- 10000000
# number of simulations
x <- 0
# number of simulations that sum = 12
for(i in 1:n_simulations){
  random_number1 <- sample(c(mat), 1, replace = TRUE)
  random_number2 <- sample(c(mat), 1, replace = TRUE)
  random_number3 <- sample(c(mat), 1, replace = TRUE)
  sum <- random_number1 + random_number2 + random_number3
  if(sum == 12){
    x <- x + 1
  }
}
Q1 <- x / n_simulations
print(Q1)

## [1] 0.1157287

Q2 <- matrix(c(200, 200, 100, 200, 200, 300, 100, 200, 100 ,100), nrow = 5, ncol = 2)
#Q2 is the form
Row5sum <- sum(Q2[5, ])
# the number of people living in 'Others'
Totalsum <- sum(Q2)
Q2 <- Row5sum / Totalsum
print(Q2)

## [1] 0.1764706

3. If the first card drawn is a diamond, then there are now 12 diamonds left out of a total of 51 cards.

Q3 <- 12 / 51
print(Q3)

## [1] 0.2352941

Q4 <- choose(20, 10)
print(Q4)

## [1] 184756

Q5 <- 20 * 20 * 18
print(Q5)

## [1] 7200

Q6 <- factorial(10) / factorial((10 - 10))
print(Q6)

## [1] 3628800

7. Event A: A coin is tossed 7 times.
   Event B: A standard six-sided die is rolled 3 times.
   Event C: A group of four cards are drawn from a standard deck of 52 cards without replacement
   

OutA <- 8
# OutA is the number of outcomes of event A.
OutB <- choose(6,3)
# OutB is the number of outcomes of event B.
OutC <- choose(52,4)
Q7 <- OutA * OutB * OutC
print(Q7)

## [1] 43316000

8. Occupying alternate seats means the sequence must be MWMWMWMW or WMWMWMWM. We need to calculate the number of one sequence and multiply by two. What’s more, this table is ROUND. If everyone move one seat to the right, the sequence is still the same. Every sequence can be moved to seven sequences, which means we need to divided by 8 at last.

PM <- factorial(4) / factorial((4 - 4))
# PM is the permutation of men
PW <- factorial(4) / factorial((4 - 4))
# PM is the permutation of women
seq <- PM * PW
Q8 <- 2 * seq / 8
print(Q8)

## [1] 144

10. Event A: withdraw one pancake and see that one side is brown.
    Event B: the other side is brown.
    So the answer should be P(B|A) = P(AB)/P(A).
    P(AB) equals to: withdraw one pancake that both sides are brown.

PAB <- 1/3
# PAB is P(AB)
PA <- 3/6
# PA is P(A). There are six sides in total, and three are brown.
Q10 <- PAB / PA
print(Q10)

## [1] 0.6666667