Probability of poker cards using combinations, its a poker question [closed]

Choose two cards from a regular 52 card deck, what is the probability of getting atleast one face card from the two card hand drawn? (Without replacement) What is the probability that the first card from the two cards you get is a facecard? (fACE CARDS ARE K,Q AND J).

Part 2 What is the probability of getting atleast one face card in your hand of 2 choosen from 52.

closed as off-topic by Grigory M, Tim Ratigan, Thomas Andrews, Adriano, PaulDec 22 '13 at 2:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Grigory M, Tim Ratigan, Thomas Andrews, Adriano, Paul
If this question can be reworded to fit the rules in the help center, please edit the question.

• so far for calculations if did 52C2 to figure out how many ways to choose 2 cards from a 52 deck, i have established there are 12 face cards. From the two cards i pick, i am not sure how to calculate that my first card will be a face and the second will not be face. Order matters. – Shawn Dec 22 '13 at 1:26

HINT: how many face-cards are there in a deck?

HINT2: what is the probability of not drawing any face card for the two cards in you hand?
The first card has probability $\frac {40}{52}$ of not being a face card, the second one $\frac{39}{51}$. The probability of at least one face card is $1-\frac{40}{52}\frac{39}{51}$. (Why?)
There are $52$ cards in the deck, with $40$ non face cards. The probability of drawing a non face card is thus $\frac {40}{52}$. now, there are $39$ non-face cards left, out of $51$, so the probability of another non-face card is $\frac{39}{51}$. The probability of two non-face cards is: $\frac{40}{52}\frac{39}{52}$. Now, drawing no face cards is the opposite of at least one face cards, so we get $1-\frac{40}{52}\frac{39}{51}$.

Does the second card influence the probabilities for the first card?
No, once the first card is drawn, you know (can calculate the probability) whether it is a face card or not. The second card isn't important anymore.

• there are 12 face cards in a deck. – Shawn Dec 22 '13 at 1:23
• 40/52 ? i am not sure – Shawn Dec 22 '13 at 1:35
• yes because for the question you only want the first card to be a facecard ....i might be wrong – Shawn Dec 22 '13 at 1:36
• Wow you totally right....i cant believe i didnt see this earlier...spent like 5 hours trying to figure this out – Shawn Dec 22 '13 at 1:44
• Can you help me with the part 2....for that part order doesnt matter – Shawn Dec 22 '13 at 1:46

Hint

What is the chance of not drawing a face card in the first draw? What is the probability of not drawing a face card in the second draw given that you did not draw a face card in the first draw? Call these $\lnot F_1$ and $\lnot F_2$, respectively. What does $1-(\lnot F_1)(\lnot F_2)$ represent?