What is the probability of drawing an ace or a club from a deck of cards?

[iceblits]

Homework Statement

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

Homework Equations

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

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}?

 

[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 I am just confused about what to do with the drawing of two cards..because drawing one card decreases the number

 

[haruspex]

I think that is my second case. AUB=A+B-A(And)B right?

Yes, so it will be easier to do b) first.

so are you saying then that a.) should be p(ace)+p(clubs)-p(ace of clubs) so (4/52)+(13/51)-(1/52)?

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.

 

[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

 

[Ray Vickson]

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 I am just confused about what to do with the drawing of two cards..because drawing one card decreases the number

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.

 

[iceblits]

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

 

[iceblits]

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.

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

 

[haruspex]

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

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?

 

[Ray Vickson]

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

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.

 

[iceblits]

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.

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

 

[iceblits]

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?

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

 

[iceblits]

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.

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

 

[iceblits]

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?

So it should be just 13/52?

 

[haruspex]

So it should be just 13/52?

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.

 

[iceblits]

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

 

[Ray Vickson]

So it should be just 13/52?

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.

 

[iceblits]

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.

So its the same...!