Probability of Two Cards Being Aces: Intro Stats Help

[dherm56]

Homework Statement

Two cards are dealt from a deck of 52 cards. Find the probability that at least one of them is an ace.

The Attempt at a Solution

This is for an intro. stat class and the material is very simple but for some reason I cannot find this answer. The only attempt that makes sense to compute is (48/52)*(4/51) but this is incorrect.

 

[Bohrok]

Not sure why you used 48 there; change it to a 4 and it should be the correct answer.

 

[dherm56]

(4/52)*(4/51) is also incorrect. I used 48 the first time b/c that is all the cards besides an ace and only one needs to be drawn

 

[Bohrok]

What is the answer?
I'm thinking you have to add the probabilities of drawing one ace and drawing two aces. Teacher didn't cover too much in probability when I studied it...

 

[dherm56]

Still no luck

 

[Bohrok]

Do you have the answer?

 

[HallsofIvy]

To get "at least one ace in two cards" you must either get an ace on the first card (in which case it doesn't matter what the second card is) or get a "non-ace" on the first card and an ace on the second card.

The probability that the first card is an ace is 4/52. The probability that the first card is NOT an ace is 48/52 and in that case, you still have 4 aces in the remaining 51 cards so, in this situation, the probabilty that the second card is an ace is 4/51. That is, the probability of "non-ace, ace" in that order is (48/52)(4/51).

The probability of "one or the other", that is the probability that at least one card of two is an ace, is the sum of those: 4/52+ (48/52)(4/51).

 

[dherm56]

That's correct, thank you very much!