Question

In the field of quality control, the science of statistics is often used to determine if a process is "out of control". Suppose the process is, indeed, out of control and 20% of items produced are defective. (a) If three items arrive off the process line in succession, what is the probability that all three are defective? (b) If four items arrive in succession, what is the probability that three are defective?

Answer 1

probability = 0.008

probability = 0.0256

Step-by-step explanation:

we know here probability of defective is 0.2

so probability of not defective is 1 - 0.2 = 0.8

as we know 3 item is arrive off process line in succession

so The probability that an item is defective is

as P(defective) = 0.20

as all item are independent so

probability that all three items are defective is

probability = 0.20 × 0.20 × 0.20 = 0.008

and

probability that exactly 3 of next 4 are defective

so number of way that can choose 3 out of 4 is

=
`(4!)/(3! ( 4-3)!)`

= 4

so as all are independent probability is

probability = ( the number of way to choose 3 out of 4 ) × ( 3 item defective ) × ( 1 item not defective )

probability =
`_4 C_3` × 0.2³ × ( 1- 0.2)

probability = 4 × 0.008 × 0.8

probability = 0.0256