Solve the Tricky Probability Problem: Drawing Cards from a Standard Deck

[davedave]

Suppose that you draw two cards without replacement from a standard deck of 52 cards.
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.

 

[Hurkyl]

How exactly are you having difficulty using conditional probabilities?

 

[daveyinaz]

Also what do you think A and B are?

 

[davedave]

Suppose that you draw two cards without replacement from a standard deck of 52 cards.
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.

i am the one that posts this problem

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?

 

[HallsofIvy]

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.

 

[spideyunlimit]

P(A l B)=P(A and B)/P(B)
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

 

[statdad]

The answer is 1/13.
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

 

[mathman]

Suppose that you draw two cards without replacement from a standard deck of 52 cards.
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.

Two cases:
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