00:01
In this problem, we have a theorem that we would like to prove.
00:02
And so in order to prove it, let us start with that left -hand side and show that it is equal to the right -hand side.
00:08
So we have the probability of a, union, b, given c.
00:13
Now, by definition, this is equal to the probability of a, union b, and c, all over the probability of c.
00:30
Now, this is equal whenever we distribute in that intersection side, of the probability of a, union c, and, and b, union, c, all over the probability of c.
00:52
Excuse me, have these signs flipped here...