Question

A company that produces fine crystal knows from experience that 12% of its goblets have cosmetic flaws and must be classified as "seconds." (Round your answers to four decimal places.)

(a)

Among nine randomly selected goblets, how likely is it that only one is a second?

(b)

Among nine randomly selected goblets, what is the probability that at least two are seconds?

(c)

If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

Answer 1

Answer:

Given,

p = 12% = 0.12

q = 1 - p

= 1 - 0.12

= 0.88

sample n = 9

consider the binomial distribution P(X = x) = nCx*p^x*q^(n-x)

a)

P(X = 1) = 9C1*0.12^1*0.88^(9-1)

= 9*0.12*0.88^8

= 0.3884

b)

To give the probability that at least two are seconds

P(X >= 2) = 1 - [P(x = 0) + P(x = 1)]

= 1 - [9C0*0.12^0*0.88^9 + 9C1*0.12^1*0.88^8]

= 1 - [0.3165 + 0.3884]

= 1 - 0.7049

= 0.2951

c)

n = 5

p = 0.88

P(4 <= x <= 5) = P(x = 4) + P(x = 5)

= 5C4*0.88^4*0.12^1 + 5C5*0.88^5*0.12^0

= 5*0.88^4*0.12 + 1*0.88^5*1

= 0.3598 + 0.5277

= 0.8875