# A basic probability question

1. Nov 29, 2012

### iceblits

1. The problem statement, all variables and given/known data
You draw 2 cards from a standard deck of cards without replacement.
a.) what is the probability that the first card is an ace OR the second card is a clubs.
b.) what is the probability that the first card is an ace AND the second card is a clubs

2. Relevant equations

3. The attempt at a solution
a.) I thought there were 2 cases to consider: your first card could be any of 3 aces that are not the club (3/52 probability) and you add to that the probability of drawing a club on your second draw (13/51). the other case is drawing the ace of clubs (probability 1/52) and adding to it the probability of drawing a club on the second try(12/51). So in all I have (3/52)+(13/51)+(1/52)+(12/51) which is a probability higher than 1/2 and it seems wrong.

b.) Same thing except this time i multiplied:
(3/52)*(13/51)+(1/52)*(12/51)

due to my suspicious answer for part (a) I suspect (b) is also wrong. I do not have the answer sheet.

-thank you

2. Nov 29, 2012

### Ray Vickson

Why can't the first card be the ace of clubs? Given two events A and B, do you recall the formula for getting P{A or B}?

3. Nov 29, 2012

### iceblits

I think that is my second case. AUB=A+B-A(And)B right? so are you saying then that a.) should be p(ace)+p(clubs)-p(ace of clubs) so (4/52)+(13/51)-(1/52)? I guess im just confused about what to do with the drawing of two cards..because drawing one card decreases the number

4. Nov 29, 2012

### haruspex

Yes, so it will be easier to do b) first.
There are two things wrong with that.
Nobody said anything about the Ace of Clubs. The Ace condition applies to one card, the Club to another.
13/51 would apply to the second card if you knew the first card was a club, but you have no information on that, whether it was Ace or not.
The events 'first card is Ace' and 'second card is club' are independent. Your original calculation for b) was correct, but it's simpler using the formula for joint probabilities of independent events.

5. Nov 29, 2012

### iceblits

Oh I see!..
So would the and question then be: P(ace first)*P(clubs on second draw|probability of ace on first draw)=4/52*12/51

edit: just realized you are referring to the (b)...I think I'm still stuck on (a)...I will think about what you said

6. Nov 29, 2012

### Ray Vickson

Sometimes it helps to think of an actual "sample space" for the experiment. In this case, imagine laying out all 52 cards in a row and then just looking at the first two cards in the row.

There are 52! possible permutations of the 52 cards. In how many of these permutations is the first card an ace? In how many permutations is the second card a club? In how many permutations is the second card the ace of clubs? From the answers to these questions you can get P{first = ace}, P{second = club} and P{second = ace of clubs}. That will let you answer (a), and you can look at (b) in a similar way.

7. Nov 29, 2012

### iceblits

Oh are you saying: AUB=A+B-A(and)B implies AUB=(4/52)+(13/51)-(the answer from part b)?

8. Nov 29, 2012

### iceblits

is it..

The first card is an ace in: 4*51 cases
The second card is a club in: 52*13 cases
The second card an ace of clubs: 52*1

9. Nov 29, 2012

### haruspex

Almost, but how do you get 13/51? Remember that B is "the second card from the deck is a club". If it asked for the probability that the last card in the deck was a club, what would you answer?

10. Nov 29, 2012

### Ray Vickson

Are you sure? Your reckoning gives
$$P\{C1 = ace\} = \frac{4 \times 51}{52!} \approx 0.253 \:10^{-65},$$ instead of the correct value 4/52 = 1/13.

11. Nov 29, 2012

### iceblits

Oh!
So it should be a factorial..i forgot about the ways to arrange the rest of the cards

12. Nov 29, 2012

### iceblits

Last card i would say...
1-(51!/(52!))?
Edit: that's not right

13. Nov 29, 2012

### iceblits

Using this method do i get:
(4*51!+13*52*51!-52*1*50!)/(52!)?

14. Nov 29, 2012

### iceblits

So it should be just 13/52?

15. Nov 29, 2012

### haruspex

Yes. The probability that any given card is a Club, with no other information, must be 1/4. The information that the first card was an ace doesn't alter this, because that card also has a 1/4 chance of being a club.

16. Nov 29, 2012

### iceblits

Oh so is it really something as simple as: 4/52+13/52-1/52 lol?

17. Nov 29, 2012

### Ray Vickson

To expand on haruspex' answer (and using the 'permutations' representation I spoke of before), the number of permutations in which the last card is a club is 13*51!, so P{last card is a club} = 13*51!/52! = 13/52 = 1/4.

18. Nov 29, 2012

### iceblits

So its the same...!

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