I'm looking for two answers for the homework

Question

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and test 49 ​tablets, then accept the whole batch if there is only one or none that​ doesn't meet the required specifications. If one shipment of 3000 aspirin tablets actually has a ​4% 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

You can think of this process as like flipping a coin. In this case, you are going to flip the coin 49 times (one for each tablet in the sample). This is not a fair coin, the probability of getting "heads" (a defective tablet) is 0.04.

You will accept the shipment if you inspect the sample if there are zero or one defective tablets. The easy way to get the probability of zero defects is to use the probability that a tablet is NOT defective, and multiply that by itself 49 times (this works if each tablet is independent of the others.)

p(1) = (1-0.04)⁴⁹ = (0.96)⁴⁹ = 0.1353

The probability of one defective in 49 tablets also follows the binomial distribution, but it's a little trickier.

p(x)= _rC_xp^x(1−p)^n−x where _rC_x is the combinatorial 49 choose 1 in this case, which equals 49.

So p(1)= ₄₉C₁(0.04)¹(1−0.04)⁴⁸ = 0.2762

Add those probabilities because the events of getting zero defective and one defective are mutually exclusive outcomes, so the probability of rejection is 0.4115. That's a lot of reject shipments, don't you think?