Question

A lot of 200 chips contain 5 that are defective. 8 are selected random with out replacement.

-The probability at least two of selected parts is not defective?
-the probability that the third part is defective given that the first and two sample is defective?

Answer 1

1 ,
`(1)/(66)` (Answer)

Explanation:

Since, 8 chips are selected at random and 5 are there defectives in the lot, So, at least (8 - 5) = 3 chosen chips will be non - defective.

So, P ( At least two of selected part is non defective ) = 1 .

P(first and second samples are defective)

=
`(5 * 4)/(200 * 199)`

P ( first, second and third samples are defective)

=
`(5 * 4 * 3)/(200 * 199 * 198)`

So,

=

=
`(3)/(198)`

=
`(1)/(66)` (Answer)