Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 55 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 5000 ​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

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.01 (1%)

N = number of trials = 55

R = desired success = at most 2

After substitution

Required probability is = P(0) + P(1) + P(2) = 0.98217123986

Yes as the probability is quite high

Almost all shipments will be accpeted