Question

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 40 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If one shipment of 5000 aspirin tablets actually has a 3​% rate of​ defects, 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 of each other

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 = 0.03

N = number of trials = 40

R = desired success = P(0) + P(1).

= 0.66154192129913279498948663112125511966004786122340140758535526194999034006060801

= 0.6615

From 100 such shipments, 66.15 ~ 66 would be accepted and 34 would be rejected.

And among 5000,

Expected value = n*p = 5000*0.6615 = 3308 would be accepted.

And 5000 - 3308 = 1692 would be rejected.