00:01
In this problem, we need to show that for any events e1 and e2, p of e1 intersection e2, is greater than or equal to p of e1 plus p of e2 minus 1.
00:11
Now for this, recall that the probability of e1 union e2 is equal to p of e1 plus p of e2 minus p of e1 intersection e2.
00:27
Now, note that p of e1 union e2, this is a probability and any probability will always be less than or equal to 1.
00:39
What does that mean? that means that this expression over here must be less than or equal to 1.
00:45
So that means that p of e1 plus p of e2 minus p of e1 intersection e2, which is equal to the probability of e1 union e2, this must be less than or equal to 1.
01:00
So let us swap both sides of this inequality.
01:05
What we end up with, we have that 1 is greater than or equal to p of e1 plus p of e2 minus p of e1 intersection e2...