Understood the problem?
A retailer buys glasses directly from the factory in large batches. The glasses come individually wrapped. Periodically inspect the retailer lots to determine the proportion of broken or chipped. If a large batch contains 10% broken or chipped, how likely the retailer to obtain a sample of 100 glasses of 17% or more lame ?
Not sure why that post was thanked. Sarcasm perhaps?
An individual pair of glasses has a probability of .1 of being defective. Out of a sample of 100 glasses, you want to know the probability that you will find 17 or more (out of the 100) glasses that are defective. Depending on what section you are in in your studies you can do this one of two ways:
Using the Binomial distribution and the complement of "17 or more defective glasses". This however involves 16 binomial calculations and (without a calculator) that is going to be quite tedious.
Or, you can use the normal as an approximation to the binomial. You know that for any one pair of glasses it has a .1 chance of being defective. Therefore you would expect there to be 10 out of the 100 to be broken, with standard deviation npq. Then you can model the binomial with the normal as N~(10, 9). At this point is is simply a matter of finding what the area to the right of 17 is on your normal curve.