AshleyT, you are right that "sampling with replacement" and "sampling with replacement" would give different results.

But this is not just a "choose with replacement" or "choose without replacement" because there is no way given to calculate a different probability given What has or has not been chosen on the first choice. The only way this problem can be done is to assume that the probabilities given apply on the second as well as the first choice no matter what happened on the first choice. That is, that the choices are independent.

If, for example, we were told that there were 1000 objects, 176 of them silver, 172 of them white, etc. Then the question of replacement would be important. The probability that the first object is silver would be 176/1000= .176. If we replace that object and choose again, the situation is exactly the same and we get the silver with probability .176 again. The probability that both are silver is . Exactly the same argument shows that the probability both objects are white is and the probability of "either 2 silver or 2 white" is the sum of those.

But if we chose a silver object first and don't replace it, we now have 999 objects, 175 of them silver. The probability of choosing a silver object the second time is now 175/999. The probability of choosing two silver objects would be (176/1000)(175/999), the probability of choosing two white objects would be (172/100)(171)/999 and the probability of choosing two silver or two white would be (176/1000)(175/999)+ (172/1000)(171/999).

But, as I said above, we are given only "probabilities", not "numbers of objects" so the only way we can do this problem is by assuming that the choices are independent.