Answer to Question #171097 in Statistics and Probability for MARCUS A MERCIER

Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 41 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 4000 ​batteries, and 2​% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?

Answer 1

Let $X=$ the number of defective batteries. The probability distribution of the random variable $X,$ hypergeometric distribution, is given by

$h(x;n,M,N)= \dfrac{\dbinom {M}{x}\dbinom{N-M}{n-x}}{\dbinom{N}{n}}$

Given that $N=4000, M=4000\cdot0.02=80,n=41.$

What is the probability that at most 3 batteries do not meet specifications?

$$P(X\leq3)=P(X=0)+P(X=1)+$$ $$+P(X=2)+P(X=3)=$$

$$={\binom{80}{0}\binom{4000-80}{41-0} \over \binom{4000}{41}} +{\binom{80}{1}\binom{4000-80}{41-1} \over \binom{4000}{41}}+$$ $$+{\binom{80}{2}\binom{4000-80}{41-2} \over \binom{4000}{41}}+{\binom{80}{3}\binom{4000-80}{41-3} \over \binom{4000}{41}}\approx$$

$\approx 0.4349+0.3676+0.1497+0.0390=0.9913$

The probability that this whole shipment will be accepted is  $0.9913.$