Probability to draw two particular cards from a deck. Given is a deck of 52 cards and the question is, what is the probability to draw an 8 and a Q (drawn without replacement).
Here is what I did:
The sample space should be $52 \choose 2$. Since for the first card, say a 8 we have 4 possibilities and for the other card we also have 4 possibilities, the probability should be :
$$
P=\frac{16}{52 \choose 2} \approx 0.01207
$$
This is unfortunately wrong, the correct answer is given approximately to be $0.0264$.
I would appreciate your help, especially an explanation why my solution is wrong.
 A: There are four Queens and four eights, so you could draw a Queen then an eight or the other way round, making the probability $$2 \times \frac4{52} \times \frac{4}{51}$$ which is your answer.
A: You are correct, and the book is wrong.   The rule of thumb for frequentistic measures is: as above, so too below. (With regards to counting combinations or permutations.)


*

*There are $\binom{4}{1}\binom{4}{1}$ ways to select 1 of 4 eights and 1 of 4 queens. 

*There are $\binom{52}{2}$ ways to select 2 of 52 cards. 

*Order is not considered in either count.   It need not be.

*If there is no bias in the selection of cards then the probability is:$$\frac{4\cdot 4\cdot 2\cdot 1}{52\cdot 51} \approx 0.012{\small (06...)}$$





*

*There are $\binom{4}{1}\cdot\binom{4}{1}\cdot 2!$ distinctly ordered ways to select 1 of 4 eights and 1 of 4 queens. 

*There are $\binom{52}{2}\cdot 2!$ distinctly ordered ways to select 2 of 52 cards. 

*Order is considered in both count.   It cancels.

*If there is no bias in the selection of cards then the probability is:$$\require{cancel}\frac{4\cdot 4\cdot 2\cdot 1\cancel{\cdot 2!}}{52\cdot 51\cancel{\cdot 2!}} \approx 0.012{\small (06...)}$$

