Question

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the possibility that (a) all 4 are defective? (b) 3 are defective and 1 is good? (c) exactly 2 are defective? (d) none are defective?

Answer 1

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Step-by-step explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a)
`C(4,4) = 1; C(15,4) = 1365`

`P = (C(4,4))/(C(15,4)) = (1)/(1365) = 0.0007326`

b)
`C(4,3) = 4; C(11,1) = 11`

`P = (C(4,3)*C(11,1))/(C(15,4)) = (4*11)/(1365) = 0.03223`

c)
`C(4,2) = 6; C(11,2) = 55`

`P = (C(4,2)*C(11,2))/(C(15,4)) = (6*55)/(1365) = 0.2418`

d)
`C(11,4) = 330`

`P = (C(11,4))/(C(15,4)) = (330)/(1365) = 0.2418`