Answer to Question #209847 in Statistics and Probability for rose

Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 50 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 3000 ​batteries, and 1​% 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 hypergeometric distribution of the random variable X:

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

N=3000

$M= 3000 \times 0.01 = 30$

n=50

The probability of whole shipment will be accepted:

$P(X≤2) = P(X=0)+P(X=1)+P(X=2) \\ P(X=0) = \frac{\binom{30}{0} \binom{3000-30}{50-0}}{\binom{3000}{50}} = 0.6024 \\ P(X=1) = \frac{\binom{30}{1} \binom{3000-30}{50-1}}{\binom{3000}{50}} = 0.3093 \\ P(X=2) = \frac{\binom{30}{2} \binom{3000-30}{50-2}}{\binom{3000}{50}} = 0.0752 \\ P(X≤2) = 0.6024 +0.3093 +0.0752= 0.9869$

The probability of whole shipment will be rejected is:

P(X>2) = 1-P(X≤2)

= 1 -0.9869

= 0.0131