Math 209 Lecture 19 - More on Probability

In this lecture, we first go over some homework solving some elementary probability problems.

We then continue our discussion of probability. We talk about independent and mutually exclusive events: how to recognize when these two kinds of events occur, and how they affect our calculations mathematically. We do a few examples to close out the lesson. We shall continue with conditional probability next.

Errata:

@53:30, 2015 Jhevon misspoke. We do NOT have independence in the no replacement case. The answer is correct though, but for a different reason.

Independence means we are allowed to use P(A and B) = P(A)*P(B). You can take A = 1st is green, and B = 2nd is green. THIS IS NOT TRUE IN THE WITHOUT REPLACEMENT CASE. HOWEVER, WHAT IS TRUE IS VERY SIMILAR.

In general, it is true that P(A and B) = P(A) * P(B | A). This works whether or not you are independent, and that is what is used here.

So, P(both Green) = P(1st is green)*P(2nd is green GIVEN that the first was Green).

It is easy to mistake this for independence, but that is not what is going on here.

Overall, you should know that we have independence in the WITH replacement case, but not in the without replacement case. Since, in the without replacement case, the sample space (and hence the probabilities) will change, so the first event has influence over the second event.

[mahmoudhassan]

Would you please explain to me why you said that two balls drawn without replacement independent and not dependent ?