Question

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?

Answer 1

There are 4 aces and there are 51 cards left after drawing a diamond.
P(ace) = 4/51

Answer 2

book's answer is 7/51

Answer 3

A standard deck of cards used for poker has only one ace in each suite, giving 4 aces. Where did the other 3 aces come from? Or, what am I missing?

Answer 4

i got 4/51 as well. couldnt figure out how the book got 7/51

Answer 5

Of course the card drawn first could be the ace of diamonds. In that case P(ace) = 3/51 :p

Answer 6

eh

Answer 7

if P(Diamond) = 13/52 (13 diamonds, 52 cards)

if only draw ace, p(ace) 1/52.

~ (1/52)/(13/52) = 1/13

Answer 8

?

Answer 9

If the ace of diamonds is drawn first then only three aces remain for the second draw.
In that case P(ace) = 3/51.
If a diamond other than the ace of diamonds is drawn first then for the second draw P(ace) = 4\51
The above events are mutually exclusive, therefore the probabilities should be added giving:\[totalP(ace)=\frac{3}{51}+\frac{4}{51}=\frac{7}{51}\]

Answer 10

thanks!

Answer 11

You're welcome :)