Statistic?

First off, being completely analytical about it, that fact that you know 2 out of 50 are defective (in a single lot) means nothing. The next lot could have 20 defective or none defective. Statistics are only really useful with big numbers. If fifty are sampled every hour for ten years and the average defective is 2, that is significant. I think that's what the question means but it doesn't say that. I'm not picking on you or the question, by the way, just showing you the level of detail you sometimes have to think with.

Assuming the question meant that 2 in 50 (on average) are defective. That means there's a 4% chance of one being defective. The question asks for the chance of at least one being defective, meaning you've got to work out the chance of one defective, two defective, three, etc. and add it all up. Or you take a short cut (which is what maths is all about). You work out the chance of none defective, knowing that the only two scenarios are none defective or at least one defective (these two chances must add up to one). Working out zero defective is easy. Chance of it being defective is 0.04 so chance of it being fine is 0.96. Chance of ten on the bounce being fine is 0.96 to the power of 10 which is 0.665. Therefore the chance of one or more being defective is 0.335 (pretty high, you notice... considering it's only a 4% defect rate). So the answer to question a is 0.335

Question b is a bit easier. What you are looking for is something called the "expectation". For instance, a coin toss is 50/50. From 10 coin tosses, how many heads do you "expect". Answer = 0.5 * 10 = 5. For our question, the probability is 0.04 and the sample size is the same (ten). Our expectation then = 0.04 * 10 = 0.4. So in every sample of ten, the average number of failures is 0.4. Or, to multiply both figures by 2.5, you expect 1 failure every 25 samples. Which fits with our initial question, which says 2 in 50.