1. Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:

a) Two hearts 13/52 *12/52 = 3/52
b) A heart on the 1st draw and a club on the 2nd draw
13/52 * 13/52 = 1/16
c) A heart on the first draw, and an ace on the second draw

I really don't know how to do c because the heart could be an ace, so I have 13/52 for the heart and then I'm not sure where to go from there. Hopefully I have the other two parts correct.

Corection Edit: For a and b I think I should have done 13/52 *12/51 = 1/17 and 13/52 * 13/51 = 13/204?

2. Originally Posted by antman
Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:
[snip]

c) A heart on the first draw, and an ace on the second draw

I really don't know how to do c because the heart could be an ace, so I have 13/52 for the heart and then I'm not sure where to go from there. Hopefully I have the other two parts correct.
One approach to see the answer might be to draw a tree diagram:

The first three branches from the first drawing are 'Ace of Hearts', 'Heart other than ace', 'No heart'.

The two branches from the second drawing are 'Ace', 'No ace'.

I get $\left(\frac{1}{52}\right) \cdot \left(\frac{3}{51}\right) + \left(\frac{12}{52}\right) \cdot \left(\frac{4}{51}\right)$.

Originally Posted by antman
Two cards are drawn successively and without replacement from an ordinary deck of cards. Find the probability of drawing:

a) Two hearts 13/52 *12/52 = 3/52
b) A heart on the 1st draw and a club on the 2nd draw
13/52 * 13/52 = 1/16

[snip]
Corection Edit: For a and b I think I should have done 13/52 *12/51 = 1/17 and 13/52 * 13/51 = 13/204?
The correction you posted is correct.