Considering Order in Probability

Question;

4 prizes allocated at random to 24 prize-winners in a lottery. Of the prizes, 4 are cars, 8 are bicycles and 12 are watches. Ann and Ben are two of the prize-winners. Show that the probability that Ann gets a car and Ben gets a bicycle or a watch is 10/69.

I understand that the answer can be obtained by:

4/24 * 20/23 = 10/69

But wouldn't the above solution fail to factor in the event where Ben's prize is first selected? I.e.:

20/24 * 4/23

The (flawed?) solution should thus be:

(4/24 * 20/23) + (20/24 * 4/23) = 20/69 ?

I get the same answer (20/69) using the selection method:

(4C1 * 20C1)/ 24C2 = 20/69

Am I doing something wrong here?

I think it works out regardless of order

If I understood your question correctly, you started with:

4/24 * 20/23 = 10/69

If you reverse the selection order, I think you get:

20/24 * 4/23, which is also 10/69, isn't it?

- Steve J