Answer to Question #103651 in Statistics and Probability for lou

Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 48 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 5000 ​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?
The probability that this whole shipment will be accepted is
nothing.
​(Round to four decimal places as​ needed.)

Answer 1

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

$$h(x;n, M, N)={\binom{M}{x}\binom{N-M}{n-x} \over \binom{N}{n}}$$

Given that $N=5000, M=5000\cdot0.02=100,n=48.$

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

$$P(X\leq3)=P(X=0)+P(X=1)+$$ $$+P(X=2)+P(X=3)=$$ $$={\binom{100}{0}\binom{5000-100}{48-0} \over \binom{5000}{48}} +{\binom{100}{1}\binom{5000-100}{48-1} \over \binom{5000}{48}}+$$ $$+{\binom{100}{2}\binom{5000-100}{48-2} \over \binom{5000}{48}}+{\binom{100}{3}\binom{5000-100}{48-3} \over \binom{5000}{48}}\approx$$ $$\approx0.377432+0.373310+0.178926+0.055379\approx$$ $$\approx0.9850$$

The probability that this whole shipment will be accepted is  $0.9850.$