# Thread: Probability(drawing cards)

1. ## Probability(drawing cards)

Just making sure I have my head on straight when it comes to the whole probability thing.

One card is drawn from a deck of 52. What is the probability that it is:
(a): A diamond?
(b):A King?
(c):An Ace or King?

Working:

(a):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, and there are 13 diamond cards so there are $_{13}C_1$ or $\frac{13!}{1!12!}=13$ ways of picking a diamond. So there the probability of drawing a diamond card is $\frac{13}{52}=\frac{1}{4}$.
(b):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, and there are 4 Kings so there are $_4K_1$ or $\frac{4!}{1!3!}=4$ ways to draw a king. So the probability of drawing a King is $\frac{4}{52}=\frac{1}{13}$.
(c):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, there are 4 Aces and 4 Kings so there are 8 cards to choose from so, there are $_8A_1$ or $\frac{8!}{1!7!}=8$ ways to draw either of the cards.
Therefore, the probability of drawing either an Ace or King is $\frac{8}{52}=\frac{2}{13}$

Am I doing it right?

2. Originally Posted by MathBlaster47
Just making sure I have my head on straight when it comes to the whole probability thing.

One card is drawn from a deck of 52. What is the probability that it is:
(a): A diamond?
(b):A King?
(c):An Ace or King?

Working:

(a):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, and there are 13 diamond cards so there are $_{13}C_1$ or $\frac{13!}{1!12!}=13$ ways of picking a diamond. So there the probability of drawing a diamond card is $\frac{13}{52}=\frac{1}{4}$.
(b):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, and there are 4 Kings so there are $_4K_1$ or $\frac{4!}{1!3!}=4$ ways to draw a king. So the probability of drawing a King is $\frac{4}{52}=\frac{1}{13}$.
(c):There are $_{52}C_1$ or $\frac{52!}{1!51!}=52$ ways of picking a card, there are 4 Aces and 4 Kings so there are 8 cards to choose from so, there are $_8A_1$ or \frac{8!}{1!7!}=8 ways to draw either of the cards.
Therefore, the probability of drawing either an Ace or King is $\frac{8}{52}=\frac{2}{13}$

Am I doing it right?
Looks good to me (apart from putting K instead of C when working out the chance of a king )

Also probability questions go here: http://www.mathhelpforum.com/math-he...s-probability/

3. Originally Posted by e^(i*pi)
Looks good to me (apart from putting K instead of C when working out the chance of a king )

Also probability questions go here: http://www.mathhelpforum.com/math-he...s-probability/
Ah.....YAY! Right answer, wrong section, got it!
Thank you!

P.S. Could I have a mod move the thread into the right place please?

4. Good job, but was putting the K instead of a C just a typo, or were you referring to the first letter of "King?" I'm just asking because I know a teacher who would take a point of for that. So, watch out!
Just trying to help

Mathemagister