Question

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

Answer 1

Answer:

1 , $\frac{1}{66}$  (Answer)

Step-by-step 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)

= $\frac{5 \times 4}{200 \times 199}$

P ( first, second and third samples are defective)

= $\frac{5 \times 4 \times 3}{200 \times 199 \times 198}$

So,

$\frac{{\textrm{Third sample is defective | first and second samples are defectives}}}$

=$\frac{{\textrm{ P ( first, second and third samples are defective)}}}{{\textrm{ P(first and second samples are defective) }}}$

= $\frac{3}{198}$

= $\frac{1}{66}$  (Answer)