1a) The triplets can be in positions 1-3 through 5-7 = 5 locations, and in 3! orders. The remaining 4 can be in 4! orders
5 * 3! * 4! = 5 * 6 * 24 = 720
1b) Find the number of ways we can have the two together, and subtract from the total
The 2 together can be in positions 1-2 through 6-7 = 6 locations, and in 2 orders. The remaining 5 can be in 5! orders.
6 * 2 * 5! = 6 * 2 * 120 = 1440
7! = 5040
5040 - 1440 = 3600
2a) There are 2 ways to roll a 2 or 5 out of 6 ways total equals 2/6 = 1/3
2b) To roll 2 5s with 2 dice, there is 1 way to roll 2 5s, and 6 * 6 = 36 possible rolls, so the probability is 1/36
3a) p = .12; n = 50
P(at least 3 defects) = 1 - P(2 or less defects)
There are 2 ways to get the binomial probability;
method 1: P(x) = C(n,x)p^x(1-p)^(n-x) = C(50,x).12^x(.88^(50-x))
P(2 or less defects) = P(0) + P(1) + P(2) = C(50,0).12^0(.88^(50-0)) + C(50,1).12^1(.88^(50-1)) + C(50,2).12^2(.88^(50-2)) = .0513
1 - .0513 = .9487
Method 2: binom.dist(2,50,.12,1) = .0513; 1 - .0513 = .9487
b) The question is whether to make a continuity correction; with such small n's, I would.
σ = √np(1-p) = √50(.12)(.88) = 2.2978
μ = np = 50 * .12 = 6
P(at least 3) = 1 - P(2 or less)
Using the continuity correction, z = (2.5-6)/2.2978 = -1.5232
1 - p(z <= -1.5232) = .9361
If we are using a table from the book which has two decimal places, our answer would be 1 - P(z <= -1.52) = .9357
4 If we are using a table that has 2 decimal places, P(63 <= x <= 73) = P((63-38)/9 <= z <= (73-38)/9) =
P(2.78 <= z <= 3.89) = .9999 - .9973 = .0026 (if we did exact calculations until round-off, we would get .0027)
