1.

part 1 --> p(1) = 3/5 = 0.6

part 2 --> p(at least 1) = 1-p(none) = 1- [ (2/5)*(1/4) ]

part 3 --> assuming that the system will be viewed as in a failed state if all 3 sampled sensors are defective --> p(3) = (3/5)*(2/4)*(1/3)

2. and 3.

use the combinations formula: nCr = n! / (r! (n-r)!)

n = number of books

r = number of spaces

4.

uses the binomial probability distribution

p(r) = nCr * p^r * q^(n-r)

part 1: you can assume that the 1 sampled part was taken directly from the process that made the parts

p = 3.5E -3 = .0035

q = 1-p = .9965

n = 1

r = 1

part 2: same assumption in part 1 is valid here as well

p = 3.5E -3 = .0035

q = 1-p = .9965

n = 3

r = 2

5.

working on it

6.

uses the normal distribution

mean = .0036

std dev = .004

x = .0045

z = (x - mean) / std dev

compute z, then use the normal distribution table to find the proportion of the area under the curve that is above z