There is a pack of $52$ cards. If we choose $5$ cards. What is the probability that the $5$ chosen cards has atleast one ace cards?

Question

There is a pack of $52$ cards. If we choose $5$ cards. What is the probability that the $5$ chosen cards has atleast one ace cards?(There are $4$ aces in a pack of $52$ cards) Well, my solution goes something like this :

If we have $4$ ace cards then the no. Of ways to choose $1$ ace card is $4C1$ then we have to choose more $4$ cards which may or may not be aces. This can be done in $51C4$ as we have $51$ cards remaining(after choosing $1$ ace card). So, the total no. Of ways to do this is $4C1×51C4$. The no. Of ways we can choose $5$ cards from $52$ cards is $52C4$ . So, the required probability is $\frac{4C1×51C4}{52C5}$.

Now if I solve this in other way:

We have $4$ ace cards on a pack of $52$ cards . If we choose $ 1 $ace card and the other non ace cards, 2 ace cards and the other non ace cards,...,4 ace cards and 1 non ace cards. These form $4$ mutually exclusive cases . So, the required probability is $\frac{4C1×48C4+4C2×48C3+4C3×48C2+4C4×48C1}{52C5}$.

However, both must yield the same answer...is the 2nd method correct?What will be the correct way?I just want to verify the solutions and also why the other one is wrong as $51C4$ includes all the cases of $1,2$ or $3$ ace cards as those $4$ cards can be anything may be ace, or a face cards or anything....

Answer 1

The second one is correct, and the first one is wrong because it overcounts (the ace of clubs, for example, may be counted in both the ace "compartment" and the other cards compartment.

But you should also know that there are simpler ways of computing P(at least one ace)

One way is to compute it using the complement,
viz compute $\;1 -$ P(no ace) $= 1 - \dfrac{^{48}C_5}{^{52}C_5}$

Answer 2

We know that in $5$ cards, one cards is guaranteed not to be an ace. We can pick this card with probability $\frac{48}{52}$.

For all the other cards to be non-aces, there are $\frac{47}{51}\frac{46}{50}\frac{45}{49}\frac{44}{48}$ ways, which gives a total probability of $\frac{48}{52}\frac{47}{51}\frac{46}{50}\frac{45}{49}\frac{44}{48}$ to not pick an ace in $5$ cards.

The probability of picking at least $1$ ace is therefore $1-\frac{48}{52}\frac{47}{51}\frac{46}{50}\frac{45}{49}\frac{44}{48}$.