I'm having trouble with this Probability mixed with Combinatorics problem. Could someone help me by explaining how should I approach these types of problems? I'm really struggling with these ones.
Five cards are chosen from a deck of 40 cards ( which corresponds to a normal deck of 52 french cards without the 8's, the 9's and the 10's ). What is the probability that you get both the ace of hearts and the ace of clubs after the extraction?
This is how I tried to solve the problem.
I wrote the probability that I get both the ace of hearts and the ace of clubs like this, by using the fact that the events are independent.
$P( Ace of Hearts and Ace of Clubs ) = P(Ace of Hearts) * P ( Ace of Clubs)$
$P(Ace OfHearts) = 1-P(NotAceOfHearts)$
$P(Ace of Clubs) = 1-P(NotAceOfClubs)$
The probability of not getting the ace of hearts can be written as the combinations without orders and without repetitions of 5 objects from 39 ( because I'm excluding the ace of hearts) over the possible combinations of 5 objects taken from 40 elements. Same thing can be done with the Ace of Clubs
$P(NotAceOfHearts) = P(NotAceOfClubs) = \frac{\binom{39}{5}}{\binom{40}{5}} = 0.875$
From these results I calculated the probability of getting the ace of hearts and the probability of getting the ace of clubs
$P(AceOfHearts) = P(AceOfClubs) = 1-P(NotAceOfHearts) = 1-P(NotAceOfClubs) = 1-0.875 = 0.125$
This implies that:
$P(AceOfHeartsandAceOfClubs) = 0.125*0.125 = 0.015625 = 1.5625\%$
The answer should be $1.282\%$
What did I do wrong?
Given that one of the cards is either desired ace, the probability that the other will not be $=\binom{38}{4}/\binom{39}{4}=0.8974$, so the probability the second desired ace appears is $0.10256$, so the final probability $=0.01282$.