Solutions HWB10

6.22: Here the random variable is binomial, with pi = .20 and n = 10, Pr(k break) = f(k) = (10 take k)(.20)^k(.80)^{10-k}

6.22a: Pr(none break) = f(0) = (.80)^10 = .1074

6.22b: Pr(two break) = f(2) = (45)(.20)^2(.80)^8 = .3020

6.22c: Pr(five break) = f(5) = (252)(.20)^5(.80)^5 = .0264

6.22d: Pr(all break) = f(10) = (.20)^10 = .0000001024 = 1.024 x 10^{-7}

6.22e: Pr(at least one breaks) = 1 = Pr(none break) = 1 – f(0) = .8926

Exercise A. You bet $1 one hundred times on a split 17-18 in roulette. This bet pays $17 for each win on a $1 bet.

Let K be the number of wins, which is binomial with n = 100, pi = 2/38.

Let W be the money won: W = 18K – 100.

mu_W = 18mu_K – 100 = 18(100)(2/38) – 100 = -5.2632 dollars.

sigma_W = 18sigma_K = 18sqrt[(100)(2/38(36/38)] = 223.2969 dollars.

You expect to lose $5.2632 give or take $223.2969, or so.

About 68% of the time your winnings will be between – $228.5601 and + $218.0337,

About 95% of the time your winnings will be in the ranbe – $5.2632 +/- $(1.96)(223.2969), and so on.