Question

Suppose that items under inspection are subject to two types of defects. About 70% of the items in a large lot to be judged to be defect free, whereas 20% have a type A defect alone and 10% have a type B defect alone. None have both types. If six of these items are randomly selected from the lot, find the probability that three have no defects, one has a type A defect and two have type B defects.

Answer: 0.042

Answer 1

This is a multinomial probability distribution.

The number of ways of partitioning n trails into 3 three different ways =

where n = y1+y2+y3

From the given data n = 6 and the trails are independent from each other.

let y1 be the event of no defect

let y2 be the event of type A defect

let y3 be the event of type B defect

y1 = 3, y2 =1, y3 =2.

P(y1=3,y2=1,y3=2) =

given p1 = 0.7, p2 = 0.2 p3 = 0.1

P(y1=3,y2=1,y3=2) = = 0.041