If you have a deck of cards and you draw 2 red cards how big is the chance of the following three being red cards

Heres another way to figure the first one.

We've have drawn 2 of 52 cards in the deck so there are 50 left.

So the total number of sets possible in picking the next three is just C(50,3).

And since we have picked two red cards, there are 24 left. And we wnat to choose any 3 of them = C(24,3)

So the probability that the next three cards are red is:    C(24,3) / C(50,3)  = 0.103265306122449

As Melody found.......

The second one is similarly figured....it would be    C(11,3) / C(50,3) = 0.0084183673469388

Again.....as Melody found......

A deck of cards has 26red and 26 black cards

2 red are gone so you now have 

24red and 26black

prob that the next 3 are red     =  $$\left({\frac{{\mathtt{24}}}{{\mathtt{50}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{23}}}{{\mathtt{49}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{22}}}{{\mathtt{48}}}}\right) = {\frac{{\mathtt{253}}}{{\mathtt{2\,450}}}} = {\mathtt{0.103\: \!265\: \!306\: \!122\: \!449}}$$  

Assoming that you pull 2 hearts instead of just two red cards what would be the probablility of the next three cards would be hearts? 

well there would be 11 hearts left and 50 cards in total.

so the prob of pulling 3 more hearts would be

$$\left({\frac{{\mathtt{11}}}{{\mathtt{50}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{10}}}{{\mathtt{49}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{48}}}}\right) = {\frac{{\mathtt{33}}}{{\mathtt{3\,920}}}} = {\mathtt{0.008\: \!418\: \!367\: \!346\: \!938\: \!8}}$$