1.  By the mailing tube, Var(I) = p(1-p), where p = P(I = 1).  But p is
the probability of at least one open space, which is 1 minus the
probability of no open spaces, 0.85^10.

2.  not graded

3.  

# remember, rgeom is "1 too small
# generate 1 random neg bin
rb <- function(r,p) sum(rgeom(r,p)) + r

ed <- function(k,b,p,nreps)
{
   sumd <- 0
   for (i in 1:nreps) {
      firstSpace <- rb(k,p)
      if (firstSpace <= b) {
         d <- b+1 - firstSpace
      } else {
         d <- firstSpace - (b+1)
      }
      sumd <- sumd + d
   }
   sumd / nreps
}

set.seed(9999);
print(ed(6,10,0.2,10000))

4.

cfd <- function(lambda) 1/sqrt(lambda)

print(cfd(3.2))


et <- function(s,nreps)
{
   sumTilFull <- 0
   for (i in 1:nreps) {
      stopNum <- 0
      numPass <- 0
      while (1) {
         stopNum <- stopNum + 1
         numAlight <- rbinom(1,numPass,0.2)
         numPass <- numPass - numAlight
         newPass <- sample(0:2,1,prob=c(0.5,0.4,0.1))
         if (numPass + newPass > s) break
         numPass <- numPass + newPass
      }
      sumTilFull <- sumTilFull + stopNum
   }
   sumTilFull / nreps

}

set.seed(9999);
print(et(5,10000))

5.