6.16: Hypergeometric. Here N = 11, N1 = 3, N0 = 8, n = 6. f(k) = (k take 3)(6-k take 8)/(6 take 11)
6.16a: Pr(0 green marbles) = f(0) = (0 take 3)(6 take 8)/(6 take 11) = 2/33 = .0609
6.16b: Pr(1 green marble) = f(1) = (1 take 3)(5 take 8)/(6 take 11) = 4/11
6.16c: Pr(1 green marbles) = f(3) = (3 take 3)(3 take 8)/(6 take 11) = 4/33
6.16d: Pr(at least one green marble) = 1 – Pr(0 green marbles) = 1 – 2/33 = 31/33
6.16e: mu = n(pi) = 6(3/11) = 18/11 =
6.16f: sigma = sqrt[(N-n)/(N-1)]sqrt[n(pi)(1-pi)] = sqrt[72]/11 = 6sqrt[2]/11 = .7714
6.20: This is binomial. pi = 3/70, n = 10, f(k) = (k take 10)(3/7)^ k(7/10)^(10-k)
6.20a: Pr(no deaths) = f(0) = (1)(1)(3/70)^10
6.20b: Pr(1 death) = f(1) = (10)(67/70)(3/70)^9
6.20c: Pr(at least 9 recover) = Pr(no more than 1 death) = (a) + (b) = (3/70)^10 + (10)(67/70)(3/70)^9
6.26: Hypergeometric. Here N = 12, N1 = 3, N0 = 9, n = 2. f(k) = (k take 3)(2-k take 9)/(2 take 12).
6.26a:
k f(k) = (k take 3)(2-k take 9)/(2 take 12).
0 12/22
1 9/22
2 1/22
6.26b:
F(x) = 0 if x < 0,
F(x) = 12/22 if 0 <= x < 1
F(x) = 21/22 if 1 <= x < 2
F(x) = 22/22 if 2 <= x
6.26c: mu = n(pi) = 2(3/12) = ½ = .5
6.26d: sigma^2 = (N – n)/(N – 1)(n)(pi)(1-pi) = (10)/(11)(2)(3/12)(9/12) = 15/44