Question

4. A bin of 52 manufactured parts contains 13 defective parts. Pick 8 parts from the bin at random (a) without replacement and (b) with replacement. In each case compute the probability that you get no defective parts.

5. A postcode consists of 5 digits of numbers from 1 to 9. Suppose a postcode is valid if it consists of at least two repeated digits (not real), e.g., 95616 and 96616 are valid but 95617 is not. Suppose one write a random postcode on a letter. What is the probability that this postcode is valid

Answer 1

4.

(a)

Number of non-defective parts = 52 - 13 = 39

Number of ways to pick 8 parts from the bin = = 752538150

Number of ways to pick 8 non-defective parts from the 39 bins = = 61523748

Probability of getting no defective parts without replacement = Number of ways to pick 8 non-defective parts from the 39 bins / Number of ways to pick 8 parts from the bin

= 61523748 / 752538150

= 0.08175499

(b)

Probability of picking defective parts = 13/52 = 0.25

Let X be the number of defective parts out of 8 picked parts.

X ~ Binomial(n = 8, p = 0.25)

Probability of getting no defective parts with replacement =

= 0.1001129

5.

Number of  5 digits post codes from numbers from 1 to 9 = = 59049

Number nonvalid post codes with all unique numbers = = 9! / (9-5)! = 9! / 4!

= 9 * 8 * 7 * 6 * 5

= 15120

Number of valid post codes = 59049 - 15120 = 43929

Probability that this postcode is valid = Number of valid post codes / Number of  5 digits post codes

= 43929 / 59049

= 0.7439415