Combined events-Card (Probability) A man draws one card at random from a complete pack of 52 playing cards, replaces it and then draws another card at random from the pack. Calculate the probability that 
i)both cards are clubs
ii)exactly one of the cards is a Queen,
iii)the two cards are identical.
My attempt, 
i)$\frac{13}{52}\cdot \frac{13}{52}=\frac{1}{16}$
How to solve the question ii and iii?
My another question is if the man draws simultaneously two cards at random, 
how the question i and ii change? Hope I can get a clear explanation. Thanks in advance.
 A: In this answer A deals with "with replacement" and B deals with "without replacement" (wich comes to the same as drawing simultaneously $2$ cards).


*

*iA) Your attempt is correct

*iB) The probability that the first card is a club is $\frac{13}{52}$. If that occurs then $51$ cards are left and $12$ of them are clubs. So the probability that the second will also be a club is $\frac{12}{51}$. That gives a probability of $\frac{13}{52}\frac{12}{51}$ that both are clubs.

*iiA) The probability that the first card is a queen is $\frac4{52}$ and the probability that the second is not a queen is $\frac{48}{52}$. That results in a probability of $\frac4{52}\frac{48}{52}$ that the first is a queen and the second is not. Likewise there is a probability $\frac{48}{52}\frac{4}{52}$ of  that the first is not a queen and the second is a queen. Then there is a probability of $\frac4{52}\frac{48}{52}+\frac{48}{52}\frac4{52}$ that exactly one of the cards is a queen.

*iiB) The probability that the first card is a queen is $\frac4{52}$. If this occurs then the probability that the second is not a queen is $\frac{48}{51}$. That results in a probability of $\frac4{52}\frac{48}{51}$ that the first is a queen and the second is not. Likewise there is a probability $\frac{48}{52}\frac{4}{51}$ that the first is not a queen and the second is a queen. Then there is a probability of $\frac4{52}\frac{48}{51}+\frac{48}{52}\frac4{51}$ that exactly one of the cards is a queen.

*iiiA) The probability that the second card will be the same as the first one is $\frac1{52}$. This because exactly $1$ of the $52$ cards that can be drawn the second time is the card that has been drawn the first time.

*iiiB) The probability that the second card will be the same as the first one is $0$. This because exactly $0$ of the $51$ cards that can be drawn the second time is the card that has been drawn the first time.

A: i) Your right on the ball here.
ii) This is slightly harder. Think of it in binary. The first one either is or isn't a queen. The same for the second.
So, assume the first will be the queen. that's 4/52 * 48/52(definitely not a queen)= 12/169. However, the same exact probability works for when the second is a queen, so we double that, or 24/169.
iii) This could be worded better. Is it that the cards have the same number and the same suit? Is it that they just have the same number?
I'll assume the second, as it's slightly harder to do. The first can be anything, and the second can be only 4/52, so the answer is 1/13.
A: (ii) "Exactly once card is queen" means you can draw either
[queen][non-queen] or [non-queen][queen]
Remember that in maths "or" means "+". 
So we have 
$$\frac{4}{52}.\frac{48}{52}  + \frac{48}{52}.\frac{4}{52}$$
(iii) "Two cards are identical" means that you draw ANY card, put it back then you draw that again. Remember probability of getting any card is 1.
$$1. \frac1{52}$$
Drawing two cards simultaneously... Remember that the first card you draw is out of 52 cards, but the second card is only out of 51 cards.
(ii) $$\frac{4}{52}.\frac{48}{51}  + \frac{48}{52}.\frac{4}{51}$$
(iii) $$1.0$$ as there are no identical cards... 
