- #1

- 50

- 0

Given that one card is a diamond, what is the probability that the other card is an ace?

How do you use the conditional probability formula P(A l B)=P(A and B)/P(B) to find the answer?

Thanks.

- Thread starter davedave
- Start date

- #1

- 50

- 0

Given that one card is a diamond, what is the probability that the other card is an ace?

How do you use the conditional probability formula P(A l B)=P(A and B)/P(B) to find the answer?

Thanks.

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

How exactly are you having difficulty using conditional probabilities?

- #3

- 225

- 0

Also what do you think A and B are?

- #4

- 50

- 0

i am the one that posts this problem

Given that one card is a diamond, what is the probability that the other card is an ace?

How do you use the conditional probability formula P(A l B)=P(A and B)/P(B) to find the answer?

Thanks.

this is how i do it using the formula

p(ace l diamond)=p(ace and diamond)/p(diamond)

"D" means diamond

p(ace and D)=p(D not ace and D ace)+p(D ace and D not ace)+p(D not ace and D not ace)

=12/52 * 1/51 + 1/52 * 12/51+12/52 * 11/51 = 1/17

p(D)=P(both D)+p(1st D and 2nd not D)+p(1st not D and 2nd D)=13/52 *12/51 +13/52 * 39/51 + 39/52 * 13/51 =15/34

so, p(ace l diamond)=2/15 This is wrong.

The CORRECT answer in the book is 7/51.

Does anyone know how to get the answer?

Last edited:

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 962

Since there are the same number of aces in each suit (1), knowing the suit of the first card does **not** give you any more information about the probability that the first card is the **ace** of diamonds or the number of aces left. The probability the second card drawn is an ace is just the same as if you did not know the suit of the first card (or if you had not drawn the first card), 4/52= 1/13.

If you want a more detailed calculation:

There are 13 diamonds in the deck so, given the the first card drawn is a diamond, the probability the**first** card drawn is the ace of diamonds is 1/13, the probability it is **not** the ace of diamonds is 12/13.

**If** the first card drawn is the ace of diamonds, then there are 3 more aces in the 51 remaining cards: the probability the second card drawn will be an ace is 3/51.

**If** the first card drawn is **not** the ace of diamonds, then there are still 4 aces in the 51 remaining cards: the probability the second card drawn will be and ace is 4/51.

Given that the first card drawn is a diamond, the probability that the second card drawn is an ace is (1/13)(3/51)+ (12/13)(4/51)= 51/(13)(51)= 1/13.

If you want a more detailed calculation:

There are 13 diamonds in the deck so, given the the first card drawn is a diamond, the probability the

Given that the first card drawn is a diamond, the probability that the second card drawn is an ace is (1/13)(3/51)+ (12/13)(4/51)= 51/(13)(51)= 1/13.

Last edited by a moderator:

- #6

- 61

- 0

where P(B) is the prob. of getting a diamond... and A is that for ace.

so P (A|B) = [ (13/52)*(4/51) + (4/52)*(13/51) ] / (13/52)

= 2*(4/51)

= 8/51

- #7

statdad

Homework Helper

- 1,495

- 35

Is is true that P(Ace | Diamond) = P(Ace and Diamond)/P(Diamond).

For the denominator: P(Diamond) = 13/52 (13 diamonds, 52 cards)

For the numerator: The ONLY way to get both an Ace and a Diamond is to draw the ace of diamonds, so the numerator is 1/52.

Answer = (1/52)/(13/52) = 1/13

- #8

mathman

Science Advisor

- 7,904

- 461

Two cases:

Given that one card is a diamond, what is the probability that the other card is an ace?

How do you use the conditional probability formula P(A l B)=P(A and B)/P(B) to find the answer?

Thanks.

Case 1. one card a diamond, one card not - probability that non-diamond is ace is 1/13.

Case 2. both cards diamonds - probability that one is the diamond ace is 2/13.

Frequency of case 1 is 2x13x39

Frequency of case 2 is 13x12

Overall probability= 17/195

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 7K

- Last Post

- Replies
- 5

- Views
- 2K

- Replies
- 6

- Views
- 3K