Suppose that a shipment contains 10 defective items and 20 n

Suppose that a shipment contains 10 defective items and 20 non defective items. If 3 items are selected at random without replacement, what is the probability that exactly 4 defective items will be obtained? The probability density function of the contamination particle size (in micrometers) can be modeled as f(x) = 2x^-3 for 1

Solution

10) total items =N= 30...defective =k=10..non-defective =N-k = 20....

n = 8..since 8 items are randomly selected!
x = 4..since out of 8 items, 4 are defective!.. n-x = (8-4) =4..

Prob. = ( 10 C 4) * ( 20 C 4) / ( 30 C 8) = 0.173836158.........

11) f(x) = 2* x^(-3) when x >1..

mean = integration from 1 to infinity [ x* 2* x^(-3) dx ] = 2....(d)

12) F(x) = integration from 1 to x [ f(x) dx ] = 1 - x^(-2)
so, F(x) = cdf = 1 - x^(-2) (b).....