# Question about conditional probability and calculating probability of drawing cards in order

You have a deck of 52 shuffled cards. You draw 5 cards randomly. 1. what is the probability the first card is a queen 2. what is the probability the second card is a queen 3. what is the probability that the first and last cards drawn are the same suit

For 1, I think the answer is 4/52. For 2, I think the answer is 4/51. For 3, I know that the sample space will have 52*51*50*49*48 different outcomes for drawing 5 cards, but I am confused about how to factor in the order of the cards drawn.

Please help! Thanks

## 1 Answer

Your answer on question 1 is correct.

Place the cards on the numbers $1,2,\dots,52$ in the understanding that the card on number $1$ will be the first card drawn, card on number $2$ the second, etc.

Has the card that was placed on e.g. number $14$ a smaller or larger probability to be a queen than the card placed on e.g. number $37$?...

No, all numbers have evidently equal chances to be "covered by a queen". That's why your answer on question 2 is wrong. Just like the card on number $1$ it has probability $\frac4{52}$ to be a queen. Your (false) answer $\frac4{51}$ is the probability that it will be a queen under the extra condition that the card on number $1$ is not a queen.

The card on number $1$ belongs to a particular suit. Then over the $51$ numbers $2,3,\dots,52$ exactly $12$ cards of that suit and $39$ cards of other suits are divided. That gives the card on number $5$ a probability of $\frac{12}{51}$ to belong also to that particular suit.

• An illuminating answer. However, you meant suit rather than suite. – N. F. Taussig Sep 18 '17 at 10:14
• @N.F.Taussig Thank you kindly. I will clear up my sins against English language (as far as I can discern them :-). Feel free to edit my answer if you still find any. – drhab Sep 18 '17 at 10:17