# Thread: Considering Order in Probability

1. ## 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?

2. ## 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

3. ## You got it!

I understand this seems like it doesn't account for a possibility but as Steve has calculated, it does account for the case. You have the right answer.