Probability question with cards

[JaMz1]

Hey,

I was stuck on this question so I'm asking you guys for some help.

"Two cards are drawn from a deck of cards without replacement"
a)What is the probability that the second card is a heart, if the first card is a heart?
b)What is the probablity that both cards are heart, given that at least one is a heart?

A) was clear for me, it was just 13-1/52-1 because you already have a heart.
But B) was not. I think you have the possible sets: If the first one is the given heartH,H) (H,C) (H,S) (H,D) and second given heart: (H,H) (C,H) (S,H) (D,H)
So the possibility of (H,H)=13*12/51*52, but we have that two times so its ~0.117 Possibilities of two different cards with one heart are: 13*13/51*52=~0.0637. This one do we have six times, so 6*0.0637+0.117=0.5. So the possibilities of (H,H) out of these sets is equal to 0.23, which is equal to 4/17. But according to my answer key(unfortunately without explanation) it should be 2/15.
So anyone who can say where I'm thinking wrong?

Many thanks

 

[Will Able]

this checks out okay

You figured the probabilities correctly, except for doubling H,H. Now you must combine the probabilities of (H,H + H,X + X,H) to give you the probability of having "at least one heart." The X indicates anything but a heart. Then your answer will be...

probability = H,H / (H,H + H,X + X,H)

H,H = (13/52)*(12/51) = ~0.5882
H,X = (13/52)*(39/51) = ~0.1912
X,H = (39/52)*(13/51) = ~0.1912

factoring out 52 & 51...
= (13*12) / ((13*12)+(13*39)+(39*13))
= 12 / (12 + 39 + 39)
= 12 / 90= 2/15 = ~0.1333