00:01
In this problem, we have in total 52 cards.
00:07
Now for the first problem, we want to find the probability that the first two cards include at least one ace.
00:15
So this is nothing but 1 minus probability that first two cards have no ace.
00:27
So this is nothing but 1 minus, so for the first two cards to have no ace, the favorable are 42, 48c2, that is from the 52 we have subtracted the 4 aces and the total possible are 52c2 and this comes as 33 over.
00:49
Next for the second one, we want the first five cards to include at least one ace.
00:55
So this is again 1 minus probability that first five have no ace.
01:05
This is nothing but 1 minus 48c5 upon 52c5.
01:14
For the third one, we want the first two cards to be pair of the same rank.
01:23
So it could be either ace -ace or 2 -2 or 3 -3 or 4 -4 and so on.
01:30
So there are 13 such pairs or 13 different values starting from 2 till ace...