Respuesta :
Answer:
1 , [tex]\frac{1}{66}[/tex] (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)
= [tex]\frac{5 \times 4}{200 \times 199}[/tex]
P ( first, second and third samples are defective)
= [tex]\frac{5 \times 4 \times 3}{200 \times 199 \times 198}[/tex]
So,
[tex]\frac{{\textrm{Third sample is defective | first and second samples are defectives}}}[/tex]
=[tex]\frac{{\textrm{ P ( first, second and third samples are defective)}}}{{\textrm{ P(first and second samples are defective) }}}[/tex]
= [tex]\frac{3}{198}[/tex]
= [tex]\frac{1}{66}[/tex] (Answer)
