Question

On an average the box contain 10 articles is likely to have 2 defectives in a consignment of 100 boxes, the total number of boxes expected to have exactly one defective are

Answer 1

Binomial distribution

$P=\frac{2}{10}=0.2, n=10$

since average $np=2$ for 10 articles.

The probability of a box having exactly one defectives is$P(x =1)$

$=\binom{10}{x}P^x(1-P)^{10-x}, x=1 \\ =\binom{10}{1}(\frac{2 }{10 })^1(\frac{8 }{10 })^{9} \\ =0.268435456$

Thus, in a consignment of 100 boxes, $0.268435456×100=26.8435456$ which is approximately 27 boxes are expected to have exactly one defectives articles.