In a 52-card deck, 3 cards are drawn. Calculate the probability of obtaining at least 1 ace.

Question

I have a problem with defining the events in probaility like for this example.

In a 52-card deck, 3 cards are drawn. Calculate the probability of obtaining at least 1 ace.

I dont understand How to seperate the events As for this example I multiplied the choice of an ace card among the four ace type cards I multiplied it by all the combinations of choosing 2 remaining cards among 51 cards and then I divided the whole by the choice of 3 cards among 52.

I would like to understand my mistake I know the solution is to use 1-p(a)

Answer 1

As pointed out in the comments, you are finding the probability of getting exactly one ace.

If we solve using your method, you need to evaluate:

P(exactly one ace) +P(exactly two aces)+P(exactly three aces) which is the same as 1-P(no ace)

Answer 2

$P(at\ least\ one\ ace)=1-P(no\ ace)=1-\dfrac{^{48}C_3}{^{52}C_3}$
In your attempt, you found out $P(exactly\ one\ ace)$. You also need to consider 2 aces, 1 other card and 3 aces.
In total you get $={^4C_1}*{^{48}C_2}+{^4C_2}*{^{48}C_1}+{^4C_3}$
Divide this by $^{52}C_3$ to get the desired probability. Actually, the compliment method is simpler.