1. P(B_1 = 1) = 0.4 2. L_1 can be 0, 1 or 2; 1 or 2 of those could alight at Stop 2, and 0, 1 or 2 could board there; so we could have as many as 4 and as few as 0; it is easily checked that all numbers in between could occur too 0:4 3. P(N = 11 | N >= 11 and N <= 20) dgeom(11-1,0.15) / (pgeom(20-1,0.15) - pgeom(10-1,0.15)) Note the -1s. 4. nreps <- 10000 nstops <- 2 suml1 <- 0 suml2 <- 0 suml1l2 <- 0 for (i in 1:nreps) { passengers <- 0 for (j in 1:nstops) { if (passengers > 0) for (k in 1:passengers) if (runif(1) < 0.2) passengers <- passengers - 1 newpass <- sample(0:2,1,prob=c(0.5,0.4,0.1)) passengers <- passengers + newpass if (j == 1) l1 <- passengers if (j == 2) l2 <- passengers } suml1 <- suml1 + l1 suml2 <- suml2 + l2 suml1l2 <- suml1l2 + l1 * l2 } print(suml1l2/nreps - (suml1/nreps) * (suml2/nreps)) # about 0.3413