Deck of playing cards

Question

Been going through an previous exam question and came across this:

5 cards are drawn from a deck of playing cards. What is the probability of drawing 3 aces? How do you calculate it using the C(n,r)?

Answer 1

So firstly there are a total of 52-choose-5 ways to draw 5 cards.

Then there are only 4-choose-3 times 48-choose-2 ways to draw 3 of the 4 possible aces along with any another 2 cards.

This gives us a probability of $$ P = \frac{\frac{4!}{3! \times 1!} \times \frac{48!}{2! \times 46!}}{\frac{52!}{4! \times 48!}} = \frac{4 \times 1128}{270725} = 0.16666 $$

using n-choose-r $=\frac{n!}{n! \times(n-r)!} $

Does this all make sense?

Answer 2

Flint72's answer has the correct logic, but a typo leading to an error. It should be:

$$P = \frac{\frac{4!}{3! \times 1!} \times \frac{48!}{2! \times 46!}}{\frac{52!}{5! \times 47!}} = \frac{4 \times 1128}{2,598,960} \approx 0.001736$$

I would have liked to post this as a comment on Flint72's answer, but I don't have enough rep to post comments yet.